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3 years ago

Please tell me the ans fast

Mathematics

2 answers:

katrin2010 [14]3 years ago

6 0

Answer:

Step-by-step explanation:

mgh=PE

(5)(9.8)(2.5)=pe

122.5j=pe

SIZIF [17.4K]3 years ago

4 0

B
is the answer to your question

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4 years ago

Read 2 more answers

An unfair coin has a probability of coming up heads of 0.65. The coin is flipped 50 times. What is the probability it will come

Lady bird [3.3K]

Answer: the probability it will come up heads 25 or fewer times is 0.019

Step-by-step explanation:

Given that;

n = 50

p = 0.65

so, q = 1 - p = 0.35

np = 50 × 0.65 = 32.5 ≥ 10

nq = 50 × 0.35 = 17.5 ≥ 10

so, we need to use Normal Approximation for the Binomial Distribution

μ = np = 50 × 0.65 = 32.5

σ = √(npq) = √( 50 × 0.65 × 0.35 ) = 3.3726

now, the probability that it will come up heads 25 or few times will be;

⇒ P( x≤25)

{using continuity correction}

⇒ P[ z < (25.5 - 32.5)/3.3726 ]

⇒ P[ z < -2.0755 ]

using z-table

= 0.01923 ≈ 0.019 { 3 decimal places}

Therefore the probability it will come up heads 25 or fewer times is 0.019

4 0

3 years ago

What proportion of US women have a height greater than 69.5 inches?

kiruha [24]

Using the Normal distribution, it is found that 0.0359 = 3.59% of US women have a height greater than 69.5 inches.

In a <em>normal distribution</em> with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

US women’s heights are normally distributed with mean 65 inches and standard deviation 2.5 inches, hence $\mu = 65, \sigma = 2.5$ .

The proportion of US women that have a height greater than 69.5 inches is <u>1 subtracted by the p-value of Z when X = 69.5</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{69.5 - 65}{2.5}$

$Z = 1.8$

$Z = 1.8$ has a p-value of 0.9641.

1 - 0.9641 = 0.0359

0.0359 = 3.59% of US women have a height greater than 69.5 inches.

You can learn more about the Normal distribution at brainly.com/question/24663213

3 0

3 years ago

Please tell me the ans fast ​

Please tell me the ans fast