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MaRussiya [10]
3 years ago
8

Help me with this please ASAP

Mathematics
1 answer:
just olya [345]3 years ago
6 0

Answer:

7) 13 in. Because the hypotenuse is the longest length than opposite and adjacent.

8) 12 in. Because it has 90° angle.

9) 12 in

10) 5 in

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What is the volume of a hemisphere with a radius of 3.8in, rounded to the nearest tenth of a cubic inch
Pavel [41]

Answer: 114.9\ in.^3

Step-by-step explanation:

Given

The radius of the hemisphere is r=3.8\ in.

The volume of a hemisphere is given by

V=\dfrac{2}{3}\pi r^3

Insert the value

V=\dfrac{2}{3}\pi \times (3.8)^3\\\\V=114.9\ in.^3

7 0
2 years ago
Suppose that scores on a knowledge test are normally distributed with a mean of 60 and a standard deviation of 4.3. Scores on an
Sphinxa [80]

Question:

(c) Boris also took a logic test. His z-score on that test was +0.93 . Does this change the answer to which test Boris performed better on? Explain your answer using z-scores.

Answer:

The answers to the questions are;

(a) Based on the z score, Boris perform better on his aptitude test.

(b) Based on the z score, Callie perform better on his knowledge test .

(c) For Boris since +0.93 = z_{logic} >  z_{knowledge}  >  z_{aptitude}

Yes as Boris now performed best on the logic test.

Step-by-step explanation:

The z-score of a score is a measurement of the score withe respect to its distance from the mean as a factor of the standard deviation.

To solve the question, we note that we are required to find the z score as follows.

z score is given by z = \frac{x -\mu}{\sigma}

Where:

z = Standard score

x = Score

σ = Standard deviation

μ = Mean

(a) To find out which test did Boris performed better usin z score, we have

Boris scored a

57 on the knowledge test and

106 on the aptitude test

Therefore the z sore for the knowledge test is

z = \frac{x -\mu}{\sigma}

Here

x = 57

μ = 60

σ = 4.3

Therefore

z_{knowledge} = \frac{57 -60}{4.3} = -3/4.3 = -30/43 = -0.6977

The z sore for Boris on the aptitude test is

Here

x = 106

μ = 110

σ = 7.1

z_{aptitude} = \frac{106 -110}{7.1} = -40/17 = -0.5634

Based on the z score, Boris perform better on the aptitude test as his z score is higher (on the number line), --0.5634, compared to the z score on the knowledge test , -0.6977

(b) For Callie we have

Callie scored a

63 on the knowledge test and

114 on the aptitude test

Therefore the z sore for the knowledge test is

z = \frac{x -\mu}{\sigma}

Here

x = 63

μ = 60

σ = 4.3

Therefore

z_{knowledge} = \frac{63 -60}{4.3} = 3/4.3 = 30/43 = 0.6977

The z sore for Callie on the aptitude test is

Here

x = 114

μ = 110

σ = 7.1

z_{aptitude} = \frac{114 -110}{7.1} = 40/17 = 0.5634

Based on the z score, Callie perform better on the knowledge test as his z score is higher (on the number line), 0.6977, compared to the z score on the aptitude test , 0.5634.  

(c) If z_{logic}  = +0.93 then sinc for Boris z_{knowledge} = -0.6977 and

z_{aptitude}=  - 0.5634 then

z_{logic} >  z_{knowledge}  >  z_{aptitude}

Therefore Boris now performed best on the logic test.

6 0
2 years ago
United Flight 15 from New York's JFK airport to San Francisco uses a Boeing 757-200 with 182 seats. Because some people with res
Tcecarenko [31]

Answer:

There is a 29.27% probability that the flight is overbooked. This is not an unusually low probability. So it does seem too high so that changes must be made to make it lower.

Step-by-step explanation:

For each passenger, there are only two outcomes possible. Either they show up for the flight, or they do not show up. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

A probability is said to be unusually low if it is lower than 5%.

For this problem, we have that:

There are 200 reservations, so n = 200.

A passenger consists in a passenger not showing up. There is a .0995 probability that a passenger with a reservation will not show up for the flight. So \pi = 0.0995.

Find the probability that when 200 reservations are accepted for United Flight 15, there are more passengers showing up than there are seats available.

X is the number of passengers that do not show up. It needs to be at least 18 for the flight not being overbooked. So we want to find P(X < 18), with \pi = 0.0995, n = 200. We can use a binomial probability calculator, and we find that:

P(X < 18) = 0.2927.

There is a 29.27% probability that the flight is overbooked. This is not an unusually low probability. So it does seem too high so that changes must be made to make it lower.

5 0
2 years ago
Theo has a balence of $-4 in his savings account. After making a deposit, he has $25 in his account.What is the overall change t
lions [1.4K]

The has a +$29 change in his account

6 0
2 years ago
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Your friend gave you a gift card for your birthday. You use it to buy a DVD for
Elza [17]
The answer is $60.00

7 0
3 years ago
Read 2 more answers
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