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The heights in inches of orangutans Are normally distributed with a population standard deviation of 3" and an unknown populatio

CaHeK987 [17]

Answer: The 95% confidence interval is approximately (55.57, 58.43)

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Explanation:

At 95% confidence, the z critical value is about z = 1.960 which you find using a table or a calculator.

The sample size is n = 17

The sample mean is xbar = 57

The population standard deviation is sigma = 3

The lower bound of the confidence interval is

L = xbar - z*sigma/sqrt(n)

L = 57 - 1.960*3/sqrt(17)

L = 55.5738905247863

L = 55.57

The upper bound is

U = xbar + z*sigma/sqrt(n)

U = 57 + 1.960*3/sqrt(17)

U = 58.4261094752137

U = 58.43

Therefore the confidence interval (L, U) turns into (55.57, 58.43) which is approximate.

7 0

3 years ago

Solve for x. 2x² +21x-61 = (x+7)² Please

Ymorist [56]

Answer:

$x = \frac{-7 +\sqrt{489} }{2}$

$x = \frac{-7 - \sqrt{489} }{2}$

Step-by-step explanation:

2x² +21x-61 = (x+7)²

expand

2x² + 21x - 61 = x² + 14x +49

move everything to one side

x² + 7x - 12 = 0

use quadratic formula

$x = \frac{-b +- \sqrt{b^{2}-4ac } }{2a}$

plug in the values

$x = \frac{-7 +- \sqrt{7^{2}-4(1)(-110) } }{2(1)}$

solve

$x = \frac{-7 +- \sqrt{49-4(1)(-110) } }{2}$

$x = \frac{-7 +- \sqrt{49 + 440 } }{2}$

$x = \frac{-7 +- \sqrt{489} }{2}$

7 0

2 years ago

Read 2 more answers

Alex is trying to run a certain number of miles by the end of the month Alex is 40% of the way to achieving her goal and she alr

docker41 [41]

Answer:

30 miles

Step-by-step explanation:

Given that:

Alex has some target to run a certain number of miles by the end of the month.

Goal already achieved = 40% of the total goal

Number of miles already run by Alex = 12 miles

To find:

Number of miles that Alex is trying to run by the end of month?

Solution:

We have to find nothing but the goal of Alex here.

Let the number of miles that Alex is trying to run by the end of the month = $x$ miles

As per question statement:

40% of total number of miles to be run = 12 miles

OR

$\dfrac{40}{100}\times x = 12\\\Rightarrow 40x=12\times 100\\\Rightarrow x = \dfrac{1200}{40}\\\Rightarrow \bold{x = 30\ miles}$

Total number of miles that Alex is trying to run by the end of the month = 30 miles

4 0

3 years ago

What multiplier do we use to increase by 72%

Deffense [45]

Answer:

1.72

Step By Step:

5 0

3 years ago

20 people applied for a job. Everyone either has a school certificate or diploma or even both. If 14 have school certificates an

STatiana [176]

Given:

Either has a school certificate or diploma or even both = 20 people

Having school certificates = 14

Having diplomas = 11

To find:

The number of people who have a school certificate only.

Solution:

Let A be the set of people who have school certificates and B be the set of people who have diplomas.

According to the given information, we have

$n(A)=14$

$n(B)=11$

$n(A\cup B)=20$

We know that,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=14+11-n(A\cap B)$

$20=25-n(A\cap B)$

Subtract both sides by 25.

$20-25=-n(A\cap B)$

$-5=-n(A\cap B)$

$5=n(A\cap B)$

We need to find the number of people who have a school certificate only, i.e. $n(A\cap B')$ .

$n(A\cap B')=n(A)-n(A\cap B)$

$n(A\cap B')=14-5$

$n(A\cap B')=9$

Therefore, 9 people have a school certificate only.

3 0

3 years ago