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1 year ago

A swimmer speeds up 1.9 m/s to 3.8 m/s during the last 11.0 seconds of the race. What is the acceleration of the swimmer?

Mathematics

1 answer:

Naddika [18.5K]1 year ago

8 0

<h2>Given :-</h2>

Initial speed (v) = 1.9m/s

Final speed (u) = 3.8m/s

Time (t) = 11 s

<h2>Formula Used :-</h2>

$v = u + a×t$ (a = acceleration)

➾ $a = {\frac{u-v}{t}}$

<h2>Solution:-</h2>

$a = {\frac{u-v}{t}}$ (putting the value of v, u and t from above)

➾ $a = {\frac{3.8-1.9}{11}}$

➾ $a = {\frac{1.9}{11}}$

➾ $a = 0.17 m/s (approx.)$

Therefore, the acceleration rate = 0.17m/s.

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3 years ago

Bill is selling stickers to raise money for a charity. He sold 36 stickers on the first day and 23 on the second day. He had 26

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He started with 92 stickers

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2 years ago

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A random sample of 150 men found that 88 of the men excercise regularly, while a random sample 200 women found that 130 of the w

melamori03 [73]

Answer:

The hypothesis is:

H₀: $p_{X}-p_{Y}=0$ .

Hₐ: $p_{X}-p_{Y}$ .

Step-by-step explanation:

Let X = number of men who exercise regularly and Y = number of women who exercise regularly.

The information provided is:

$n_{X}=150\\X=88\\n_{Y}=200\\Y=130$

Compute the sample proportion of men and women who exercise regularly as follows:

$\hat p_{X}=\frac{X}{n_{X}}=\frac{88}{150}=0.587$

$\hat p_{Y}=\frac{Y}{n_{Y}}=\frac{130}{200}=0.65$

The random variable X follows a Binomial distribution with parameters n = 150 and $\hat p_{X}=0.587$ .

The random variable Y also follows a Binomial distribution with parameters n = 200 and $\hat p_{Y}=0.65$ .

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\hat p=p$

The standard deviation of this sampling distribution of sample proportion is:

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

So, the sampling distribution of the proportion of men and women who exercise regularly follows a Normal distribution.

A two proportion z-test cab be performed to determine whether the proportion of women is more than men who exercise regularly.

The hypothesis for this test cab be defined as:

H₀: The proportion of women is same as men who exercise regularly, i.e. $p_{X}-p_{Y}=0$ .

Hₐ: The proportion of women is more than men who exercise regularly, i.e. $p_{X}-p_{Y}$ .

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3 years ago

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in the unted states, the height of men are normally distributed with the mean 69 inches and standard deviation 2.8 inches. If 16

yaroslaw [1]

Answer:

Probability that their mean height is less than 68 inches is 0.0764.

Step-by-step explanation:

We are given that in the united states, the height of men are normally distributed with the mean 69 inches and standard deviation 2.8 inches.

Also, 16 men are randomly selected.

Let $\bar X$ = sample mean height

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean height = 69 inches

$\sigma$ = population standard deviation = 2.8 inches

n = sample of men = 16

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the mean height of 16 randomly selected men is less than 68 inches is given by = P( $\bar X$ < 68 inches)

P( $\bar X$ < 68 inches) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{68-69}{\frac{2.8}{\sqrt{16} } }$ ) = P(Z < -1.43) = 1 - P(Z $\leq$ 1.43)

= 1 - 0.9236 = 0.0764

Now, in the z table the P(Z x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.43 in the z table which has an area of 0.92364.

Therefore, probability that their mean height is less than 68 inches is 0.0764.

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3 years ago

Elixir of acetaminophen contains 160 milligrams per 5 milliliters. How much acetaminophen is needed to prepare 2 ounces of elixi

frutty [35]

1 UK fl oz = 28,413 ml
2 UK fl oz = 56,826 ml

If 5ml contains 160mg
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________________________

56,826 X 160 = 9,092,160

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3 years ago

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