Classify the following triangle .check all that apply

Answer:

Protista

Step-by-step explanation:

Archaebacteria.

Eubacteria.

Protista.

Fungi.

Plantae.

Animalia.

These are the 6 kingdoms

Answer:

/-37/, /-24/, /-6/, /2/, /18/, /23/

The PIN is 4829

Step-by-step explanation:

let s take 4 numbers a b c and d

the PIN is abcd

we know that

(1) a = b/2

(2) b+c=10

(3) d=b+1

(4) a+b+c+d=23

from (2) c = 10 - b

from (3) d = b + 1

so (4) gives

b/2 + b + 10 - b + b +1 = 23

so

3/2 b = 23 -11 = 12

b = 12*2/3 = 8

so d = 9

c = 10-8=2

and a = 4

so the PIN is 4829

thank you

Answer:

a) The standard deviation of the annual income σₓ = 2045

b)

The calculated value Z = 0.608 < 1.645 at 10 % level of significance

Null hypothesis is accepted

The average annual income is greater than $32,000

c)

The calculated value Z = 1.0977 < 1.96 at 5 % level of significance

Null hypothesis is accepted

The average annual income is  equal to  $33,000

d)

95% of confidence intervals of the Average annual income

(26 ,746.8 ,34, 763.2)

Step-by-step explanation:

Given size of the sample 'n' =100

mean of the sample x⁻ =  $30,755

The Standard deviation = $20,450

a)

The standard deviation of the annual income σₓ = [tex]\frac{S.D}{\sqrt{n} }[/tex]

                                               = [tex]\frac{20,450}{\sqrt{100} }= 2045[/tex]

b)

Given mean of the Population μ =  $32,000

Given size of the sample 'n' =100

mean of the sample x⁻ =  $30,755

The Standard deviation ( σ)= $20,450

Null Hypothesis:- H₀: μ > $32,000

Alternative Hypothesis:H₁: μ <  $32,000

Level of significance α = 0.10

[tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]

[tex]Z = \frac{30755-32000 }{\frac{20450}{\sqrt{100} } }[/tex]

Z= |-0.608| = 0.608

The calculated value Z = 0.608 < 1.645 at 10 % level of significance

Null hypothesis is accepted

The average annual income is  greater than $32,000

c)

Given mean of the Population μ =  $33,000

Given size of the sample 'n' =100

mean of the sample x⁻ =  $30,755

The Standard deviation ( σ)= $20,450

Null Hypothesis:- H₀: μ =  $33,000

Alternative Hypothesis:H₁: μ ≠ $33,000

Level of significance α = 0.05

[tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]

[tex]Z = \frac{30755-33000 }{\frac{20450}{\sqrt{100} } }[/tex]

Z = -1.0977

|Z|= |-1.0977| = 1.0977

The 95% of z -value = 1.96

The calculated value Z = 1.0977 < 1.96 at 5 % level of significance

Null hypothesis is accepted

The average annual income is equal to  $33,000

d)

95% of confidence intervals is determined by

[tex](x^{-} - 1.96 \frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} })[/tex]

[tex](30755 - 1.96 \frac{20450}{\sqrt{100} } , 30755 +1.96 \frac{20450}{\sqrt{100} })[/tex]

( 30 755 - 4008.2 , 30 755 +4008.2)

95% of confidence intervals of the Average annual income

(26 ,746.8 ,34, 763.2)

Answer:

Dear Laura Ramirez

Answer to your query is provided below

Option D is correct.

Reason - Because of Hinge and Converse of Hinge theorem

Answer:

[tex]\left ( 8\times 2 \right )-\left ( 3\times 4 \right )[/tex]

Step-by-step explanation:

Given: The statement is ' the product of eight  and two, minus the product of three and four'

To find: expression for the given statement

Solution:

An algebraic expression is an expression consists of coefficients, variables, and the arithmetic operations.

Product of eight  and two = [tex]\left ( 8\times 2 \right )[/tex]

Product of three and four = [tex]\left ( 3\times 4 \right )[/tex]

Therefore,

Product of eight  and two, minus the product of three and four = [tex]\left ( 8\times 2 \right )-\left ( 3\times 4 \right )[/tex]

Answer:

x = 12

Step-by-step explanation:

g(x)= 2x-4

g(x)= 20

Therefore,

2x-4 = 20

Bringing -4 to the other side it becomes positive,so..

2x= 20+4

= 24

x =24/2

= 12

Answer:

= 19300

Step-by-step explanation:

Each claim consists of two parts = X + Y

where

X = the benefit that is paid to the surgeon and

Y = benefit that is paid to the hospital

V(X) = 5000, V(Y) = 10000 and V(X+Y) = 17000

So V(X+Y) = V(X) + V(Y) + 2cov(X,Y)

17000 = 5000 + 10000 +2 cov(X,Y)

17000 -15000 = 2cov(X,Y)

2000 = 2cov(X,Y)

cov(X,Y) = 1000

Now X is increased by flat Rs. 100 per claim and Y by 10% per claim

total benefit = X+100+Y+0.1Y = X+100 + 1.1Y

V(total benefit) = V(X) + 1.1²V(Y) +2(1.1)cov(X,Y) [ V(aX+bY)

= a²V(X) +b²V(Y) +2abcov(X,Y) and V(X+c) = V(X)]

= 5000 + (1.21*10000) + (2.2*1000)

= 5000 + 12100 + 2200

= 19300

Answer:

65.1 and 69.1

Step-by-step explanation:

a^2+b^2=c^2

c=95

b=a+4

Solve for a^2+(a+4)^2=95^2

a=65.1

b=a+4=69.1

Answer:

65.1 and 69.1

Step-by-step explanation:

c² = a² + b²

c= 95

a - one leg

b= (a + 4) - second leg

95² = a² + (a + 4)²

9025 = a² + a² + 2*4a + 16

2a² + 8a - 9009 = 0

[tex]a= \frac{-b +/-\sqrt{b^2 - 4ac} }{2a} \\\\a = \frac{-8 +/-\sqrt{8^2 - 4*2*9009} }{2*2} \\\\a=65.1 \ and \ a=- 69.1[/tex]

A leg length can be only positive. a = 65.1

b = 65.1 + 4 = 69.1

Answer:

The capital B refers to the base of the area

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

The capital B means the area of the base

Step-by-step explanation:

178.25

The number 5 is in the place of one's so the value of 5 is 5

Answer:

Step-by-step explanation:

  We know that = Liability

[tex]PLiability= \frac{6000}{1.05^{4} }[/tex]

[tex]\frac{6000}{1.05^{4} }=\frac{A}{1.05^{2} }+\frac{B}{1.05^{6} }\\\\6000(1.05^{2} ) = (1.05^{4} ) +B\\B= 6000(1.05^{2} )-(1.05^{4} )----------(1)\\\\[/tex]

dAssets =dLiability

[tex]4=2*\frac{\frac{A}{1.05^2} }{\frac{6000}{1.05^4} } +6*\frac{\frac{B}{1.05^6} }{\frac{6000}{1.05^4} } \\4={\frac{6000}{1.05^4}= 2*\frac{A}{1.05^2} +6*\frac{B}{1.05^6}\\\\4[6000(1.05^2)]= 2*A(1.05^4)+6*B[/tex]

From equation 1 we have

[tex]4[6000(1.05^2)]= 2*A(1.05^4)+6*6000(1.05^2)-6*A(1.05^4)\\4*A(1.05^4)=2*6000(1.05^2)\\A=\frac{2*6000(1.05^2)}{4*(1.05^4)} \\A=272.088435\\[/tex]

Going back to equation 1 we have

[tex]B= 6000(1.05^2)-A(1.05^4)\\B= 3307.5\\|A-B|= |2721.088435-3307.5|= 586.411565[/tex]

Answer:

The answer is A

Step-by-step explanation:

Answer:

13 mins

Step-by-step explanation:

8.03- 1.85= 6.18

-.72=5.46

/.42=13

Answer: Biased sample

Step-by-step explanation:

This is a biased sample because only students with strong opinions are likely going to volunteer or show interest in representing the school at the board meeting. This sample is a voluntary type sample, and at such the conclusion is not valid. This sample is biased because a group or population of students have a higher or lower sampling probability.

Answer:

60%

Step-by-step explanation:

Percent means out of 100

Changing 3/5 to a denominator of 100

3/5*20/20

60/100

The percent is 60 %

Answer:

Option B.

Step-by-step explanation:

Option A.

f(x) = [tex](0.5)^{x}[/tex]

Derivative of the given function,

f'(x) = [tex]\frac{d}{dx}(0.5)^x[/tex]

      = [tex](0.5)^x[\text{ln}(0.5)][/tex]

      = [tex]-(0.693)(0.5)^{x}[/tex]

Since derivative of the function is negative, the given function is decreasing.

Option B. f(x) = [tex]5^x[/tex]

f'(x) = [tex]\frac{d}{dx}(5)^x[/tex]

      = [tex](5)^x[\text{ln}(5)][/tex]

      = [tex]1.609(5)^x[/tex]

Since derivative is positive, given function is increasing.

Option C. f(x) = [tex](\frac{1}{5})^x[/tex]

f'(x) = [tex]\frac{d}{dx}(\frac{1}{5})^x[/tex]

      = [tex]\frac{d}{dx}(5)^{(-x)}[/tex]

      = [tex]-5^{-x}.\text{ln}(5)[/tex]

Since derivative is negative, given function is decreasing.

Option D. f(x) = [tex](\frac{1}{15})^x[/tex]

                f'(x) = [tex]-15^{-x}[\text{ln}(15)][/tex]

                       = [tex]-2.708(15)^{-x}[/tex]

Since derivative is negative, given function is decreasing.

Option (B) is the answer.

Answer:

= 0.0041

Step-by-step explanation:

Given that:

A farmer was interest in determining how many grasshoppers were in his field. He knows that the distribution of grasshoppers may not be normally distributed in his field due to growing conditions. As he drives his tractor down each row he counts how many grasshoppers he sees flying away

mean number of flights to be 57

a standard deviation of 12

fewer flights on average in the next 40 rows

[tex]\mu = 57\\\\\sigma=12\\\\n=40[/tex]

so,

[tex]P(x<52)[/tex]

[tex]=P(\frac{x-\mu}{\sigma/\sqrt{n} } <\frac{52-57}{12/\sqrt{40} } )\\\\=P(z<\frac{-5\times6.325}{12} )\\\\=P(z<\frac{-31.625}{12})\\\\=P(z<-2.64)[/tex]

using z table

= 0.0041

The probability of the farmer will count 52 or fewer flights on average in the next 40 rows down which he drives his tractor is 0.0041 and this can be determined by using the properties of probability.

Given :

The probability of the farmer will count 52 or fewer flights on average in the next 40 rows down which he drives his tractor, can be determined by using the following calculations:

[tex]\rm P(x<52)=P\left (\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n} }}<\dfrac{52-57}{\dfrac{12}{\sqrt{40} }}\right)[/tex]

[tex]\rm P(x<52)=P\left (z<\dfrac{-5\times 6.325}{12 }}\right)[/tex]

[tex]\rm P(x<52)=P\left (z<\dfrac{-31.625}{12 }}\right)[/tex]

[tex]\rm P(x<52)=P\left (z<-2.64\right)[/tex]

Now, using z-table:

P(x < 52) = 0.0041

For more information, refer to the link given below:

https://brainly.com/question/21586810

The open circle means we do not include the endpoint, hence the use of a greater than symbol. If we were to include the endpoint, then we'd have greater than or equal to. We can rule out choice B due to this reasoning.

The ray pointing to the right indicates we are talking about x values larger than 5, so we can rule out choice A and conclude the answer is C.

Side note: The notation [tex]x \in \mathbb{R}[/tex] is saying "x is a real number"

Answer:

A.

Step-by-step explanation:

Volume of sphere = (4/3)πr³

= 4/3(3.14)(d/2)³

= 4/3(3.14)(7/2)³

= 4/3(3.14)(3.5)³

= 4/3(3.14)(42.875)

= (1.33)(3.14)(42.875)

= 179.5 cm³

Answer:

The answer is 179.5 cm^2

Answer:

the dimension of the poster = 90 cm length and 60 cm  width i.e 90 cm by 60 cm.

Step-by-step explanation:

From the given question.

Let p be the length of the of the printed material

Let q be the width of the of the printed material

Therefore pq = 2400 cm ²

q = [tex]\dfrac{2400 \ cm^2}{p}[/tex]

To find the dimensions of the poster; we have:

the length of the poster to be p+30 and the width to be [tex]\dfrac{2400 \ cm^2}{p} + 20[/tex]

The area of the printed material can now be:  [tex]A = (p+30)(\dfrac{2400 }{p} + 20)[/tex]

=[tex]2400 +20 p +\dfrac{72000}{p}+600[/tex]

Let differentiate with respect to p; we have

[tex]\dfrac{dA}{dp}= 20 - \dfrac{72000}{p^3}[/tex]

Also;

[tex]\dfrac{d^2A}{dp^2}= \dfrac{144000}{p^3}[/tex]

For the smallest area [tex]\dfrac{dA}{dp }=0[/tex]

[tex]20 - \dfrac{72000}{p^2}=0[/tex]

[tex]p^2 = \dfrac{72000}{20}[/tex]

p² = 3600

p =√3600

p = 60

Since p = 60 ; replace p = 60 in the expression  q = [tex]\dfrac{2400 \ cm^2}{p}[/tex]   to solve for q;

q = [tex]\dfrac{2400 \ cm^2}{p}[/tex]

q = [tex]\dfrac{2400 \ cm^2}{60}[/tex]

q = 40

Thus; the printed material has the length of 60 cm and the width of 40cm

the length of the poster = p+30 = 60 +30 = 90 cm

the width of the poster = [tex]\dfrac{2400 \ cm^2}{p} + 20[/tex] = [tex]\dfrac{2400 \ cm^2}{60} + 20[/tex]  = 40 + 20 = 60

Hence; the dimension of the poster = 90 cm length and 60 cm  width i.e 90 cm by 60 cm.

Answer:

C.

Step-by-step explanation:

[tex]\frac{1}{5} +\frac{5}{6}[/tex] ≈ 1

5 + 8 + 1 = 14

Answer:

[tex]5(3x-2y-2)[/tex] i think. i am sorry if i am wrong

Step-by-step explanation:

Answer:

630 milliliters.

Step-by-step explanation:

The statement tells us that the final product is 14 jars of honey peanut butter and that in each jar use 45 mliliters of honey. This means that to know the total honey to be used, the required quantity for each jar must be multiplied by the total number of jars, that is:

14 * 45 = 630

Which means that he would spend a total of 630 milliliters.

Step-by-step explanation:

Ares uses 630 ml of honey, because 14×45=630

Answer:

6.75 ft

Step-by-step explanation:

81 inches

We know there are 12 inches in 1 ft

81 inches * 1 ft/ 12 inches = 81/12 ft =6.75 ft

Answer:

Step-by-step explanation:

Answer:

5/6

Step-by-step explanation:

Find the factor that divides both numbers...

25/5=5

30/5=6

5/6 is the simplified ratio

P.S. Please give me brainliest, i have only have two!

Answer:

1/3:1/4

Step-by-step explanation:

203/259

write in simplest form

2 hours to 45 seconds

Express ratio

15:1

simplest form

1/3:1/4

Answer:

Part a

For this case n = 4. If we use the future value formula we got:

[tex] A= 2500 (1+ \frac{0.0675}{4})^{4*10}= 4882.506[/tex]

Part b

For this case n = 365. If we use the future value formula we got:

[tex] A= 2500 (1+ \frac{0.0675}{365})^{365*10}= 4909.776[/tex]

Step-by-step explanation:

We can use the future vaue formula for compound interest given by:

[tex] A= P(1+ \frac{r}{n})^{nt}[/tex]

Where P represent the present value, r=0.0675 , n is the number of times that the interest is compounded in a year and t the number of years.

Part a

For this case n = 4. If we use the future value formula we got:

[tex] A= 2500 (1+ \frac{0.0675}{4})^{4*10}= 4882.506[/tex]

Part b

For this case n = 365. If we use the future value formula we got:

[tex] A= 2500 (1+ \frac{0.0675}{365})^{365*10}= 4909.776[/tex]

Answer:

f(g(x))=12x+1

Step-by-step explanation:

f(g(x)) = -2(-6x+3)+7

f(g(x))= 12x-6+7

f(g(x))=12x+1

Domain: All real numbers

Answer:

The data support the hypothesis that the proportion of defectives is lower for robots than humans.

Step-by-step explanation:

To know if the proportion of defectives is lower for robots than humans so as to prove if the hypothesis is true.

From the data given:

Total number of cables a robot assembled = 507

Defectives = 9

To get the defect rate =  the number of defects divided by the total number of cables, multiplied by 100.

Defect rate =  (9 / 507) x 100 = 0.01775 x 100

Defect rate for the robot = 1.775‬%

From the question, a robot was used and the defect rate after the calculation is 1.775‬%. While for humans, the defect rate is 3.5%. This implies, if humans were used to assembling the same 507 cables, there will be 17.745‬ defectives.

x / 507 = 3.5%

x (defectives) = 17.745‬

Therefore, the data support the hypothesis that the proportion of defectives is lower for robots than humans.