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2 years ago

HELP! Evaluate the expression below for s = 4 and t = 6. 4st² – s²

Mathematics

2 answers:

Leto [7]2 years ago

6 0

Answer:

560

Step-by-step explanation:

joja [24]2 years ago

6 0

Answer:

560

Step-by-step explanation:

substitute "s" and "t" into the equation:

4(4)(6)^2-(4)^2

when you solve this according to the order of operations, you will get 560

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More free points cause I’m the goat ya dig

denis23 [38]

Answer:

YO! THANKS!

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

As people exited a movie theater, they were informally surveyed about whether they enjoyed the movie or not.

Ilia_Sergeevich [38]

A=41% B=3% hope this helps

4 0

2 years ago

1.Famous basketball player Michael Jordan had a free throw percentage of 0.83. (He made 83% of his free throw shots.) Assume tha

MissTica

Michael Jordan's Freethrow, the population of the 6 countries and the distribution of the uninsured drivers are illustrations of probabilities

<h3>1. Michael Jordan Freethrow</h3>

The given parameter is:

p = 83% -- the percentage of free throws he made

<u>Probability that he misses his 6th throw </u>

The probability that he misses his 6th throw is:

P = P(He made the first 5 throws) * P(He lost the 6th)

This gives

P = (83%)^5 * (1 - 83%)^1

P = 0.0670

<u>Expected number of miss</u>

The expected number till he misses the first free throw is:

E(x) = np

This gives

E(x) = 6 * 83%

E(x) = 4.98

Approximate

E(x) = 5

Hence, the probability that he misses his 6th throw is 0.0670 and the expected number till he misses the first free throw is 5

<h3>2. Country Populations</h3>

The given parameters are:

Average population = 268.35 million

Standard deviation = 370.43 million

<u>The z-scores for the United States, Egypt, and Mexico</u>

This is calculated as:

z = (x - $\mu$ )/ $\sigma$

So, we have:

US = (332 million - 268.35 million)/ 370.43 million

US = 0.17

Egypt = (104 million - 268.35 million)/ 370.43 million

Egypt = -0.44

Mexico = (128 million - 268.35 million)/ 370.43 million

Mexico = -0.38

<u>The countries from the smallest to the largest </u>

By comparing the z-score of the countries, the order of the countries are:

Egypt, Japan, Mexico, Brazil, United States and India

The countries are ranked that way because the higher the z-score, the higher the population

<h3>3. Uninsured drivers</h3>

The given parameter is:

p = 12.6% -- the percentage of uninsured drivers

<u>Probability that exactly 5 drivers of 30 are uninsured</u>

This is calculated as:

P(x) = nCx * p^x * (1 - p)^(n-x)

So, we have:

P(5) = 30C5 * (12.6%)^5 * (1 - 12.6%)^(30-5)

Evaluate the product

P(5) = 0.1561

<u>Probability that the next uninsured driver is the 7th</u>

This is calculated as:

P = P(First 6 are insured) * P(7th is unisured)

This gives

P = (1 - 12.6%)^6 * (12.6%)^1

P = 0.0562

The probability that exactly 5 drivers of 30 are uninsured is 0.1561, and the probability that the next uninsured driver is the 7th is 0.0562

Read more about probability at:

brainly.com/question/25870256

3 0

2 years ago

PLEASE HELP!!!!!!!!!!

kobusy [5.1K]

73 feet per second

Steps-
Miles to cover- 50 miles = 50*5280 feet = 264,000 feet
Time it takes- 1 hour = 1*60*60 seconds = 3,600 seconds

So speed is- 264,000 feet per 3,600 seconds = 264,000 / 3,600 = 73.33 ft per second

4 0

2 years ago

After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni

antoniya [11.8K]

Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) $p_v =P(z>2.8)=1-P(z$

c) Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=64 represent were in favor of firing the coach

$\hat p=\frac{64}{100}=0.64$ estimated proportion for were in favor of firing the coach

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of Americans for were in favor of firing the coach

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561$

$0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719$

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =100*0.64=64>10$

$n(1-p_o)=100*(1-0.64)=36>10$

Calculate the statistic

The statistic is calculated with the following formula:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}$

On this case the value of $p_o=0.5$ is the value that we are testing and n = 100.

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8$

The p value for the test would be:

$p_v =P(z>2.8)=1-P(z$

Part c

Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

6 0

3 years ago