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4 years ago

PLEASE HELP ME!!! PLEASE TRY TO DO ALL OF THEM THX! :)

Mathematics

1 answer:

kakasveta [241]4 years ago

5 0

Hello there!

Number 1 is C (11, 4) Image is below

I don't know number 2, so so sorry!

Number 3 is A. 36 The formula for perimeter of a rectangle is (2 * base) + (2 * height)
The base = 13
The height = 5

(5 * 2) + (13 * 2)

10 + 26 = 36

Number 4 is 23 units.

I hope I was of assistance!

~ Fire

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Of the entering class at a college, 75% attended public high school, 15% attended private high school, and 10% were home s

poizon [28]

Answer:

a)- 0.7670

b)- 0.176

c)- 0.809

Step-by-step explanation:

Let us assume that,

P(A)= Attended Public School

=0.75

P(B)= Attended Private School

=0.15

P(C)= Attended Home School

=0.10

P(E)= Student made the Dean's list

Then,

P(E/A)= Probability of Student made the Dean's list given that he\she attended Public School = 0.18

P(E/B)= Probability of Student made the Dean's list given that he\she attended Private School = 0.16

P(E/C)= Probability of Student made the Dean's list given that he\she is from Home School = 0.17

a)- P(E)= Probability that Student made the Dean's List

= P(A) × P(E/A) + P(B) × P(E/B) + P(C) × P(E/C)

= 0.75 × 0.18 + 0.15 × 0.16 + 0.10 × 0.17

= 0.176

b)- P(A/E) = Probability that Student came from a Public high school, given that Student made the Dean's list

= [ P(E/A) × P(A) ] ÷ P(E)

= [ 0.18 × 0.75 ] ÷ 0.176

= 0.767

c)- P(C'/E') = Probability that Student was not home schooled, given that the Student did not make the Dean's list

= [ P(E'/C') × P(C') ] ÷ P(E')

= [ 0.741 × 0.90 ] ÷ 0.82 [∵P(E'/C')=P(A)×P(E'/A)+P(B)×P(E'/B)

= 0.75 × 0.82 + 0.15 × 0.84

= 0.741

= 0.809

8 0

4 years ago

G(t)=2+5<br>f(t) = -1² + 5<br>Find (8 + f)(t)

jolli1 [7]

Answer:

Step-by-step explanation:

Given that,

G(t) = 2+5

F(t) = -1^2 + 5

= (-1 x -1) = 1

Therefore, F(t) = 1 + 5

G(t)= 7

Substitute 7 for(t)

F(7) =1+5

To find (8+f)(t)

Substitute all for t

(8+6)x7

=(14)x7

=98

5 0

3 years ago

The weights of adult male beagles are normally distributed, with a mean of 25 pounds and a standard deviation of 4 pounds.

Andrews [41]

Answer:

A) 8

Step-by-step explanation:

Mean weight = u = 25 pounds

Standard Deviation = $\sigma$ = 4 pounds

We have to find how many beagles out of 75 will weigh more than 30. Since the data is normally distributed, we can use z score to find this value. First we will what is the percentage(probability) of a randomly selected beagle to weigh more than x = 30 pounds, using this percentage we can then find our answer.

The formula for z score is:

$z=\frac{x-u}{\sigma}$

Using the values, we get:

$z=\frac{30-25}{4}=1.25$

So, P(Weight > 30) is equivalent to P(z > 1.25). Using the z table, we can write:

P(z > 1.25) = P(Weight > 30) = 0.1056

This, 0.1056 or 10.56% of the beagles are expected to weigh more than 30 pounds.

So,out of 75 beagles, 10.56% of 75 are expected to weigh more than 30 pounds.

10.56% of 75 = 0.1056 x 75 = 7.92 = 8 (rounding to nearest integer)

Therefore, out of 75 beagles 8 are expected to weigh more than 30 pounds.

6 0

3 years ago

The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?

Serhud [2]

Answer:

$b = \sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}}$

Step-by-step explanation:

Given - The circumference of the ellipse approximated by $C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }$ where 2a and 2b are the lengths of 2 the axes of the ellipse.

To find - Which equation is the result of solving the formula of the circumference for b ?

Solution -

$C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }\\\frac{C}{2\pi } = \sqrt{\frac{a^{2} + b^{2} }{2} }$

Squaring Both sides, we get

$[\frac{C}{2\pi }]^{2} = [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2} } = {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2} } = {{a^{2} + b^{2} }$