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3 years ago

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Chuck drops a rock from a height of 70 m. How long will it take the rock to hit the ground? Use the formula h(t)=−4.9 t 2 + v o

t+ h o , where v o is the initial velocity and h o is the initial height. Round to the nearest tenth of a second.

2 answers:

Pachacha [2.7K]3 years ago

7 0

That's the best I could do. I hope it helped

storchak [24]3 years ago

7 0

Answer: 3.8

Step-by-step explanation:

your suppose to round (btw do u have the rest of this test answers yet because i need them too)

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Multiples between 61 and 107 that is a multiple of 4,10 and 20

Vilka [71]

80 might be the answer you are looking for

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4 years ago

Based on historical data, your manager believes that 26% of the company's orders come from first-time customers. A random sample

scoundrel [369]

Answer:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$

Step-by-step explanation:

For this case we have the following info given:

$p = 0.26$ represent the proportion of the company's orders come from first-time customers

$n=158$ represent the sample size

And we want to find the following probability:

$p(\hat p >0.4)$

And we can use the normal approximation since we have the following two conditions:

1) np = 158*0.26 = 41.08>10

2) n(1-p) = 158*(1-0.26) = 116.92>10

And for this case the distribution for the sample proportion is given by:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$

8 0

3 years ago

A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are list

Kitty [74]

Answer:

It does not appear that, as a group, the students are reasonably good at estimating one minute.

Step-by-step explanation:

We are given the following data in the question:

75, 88, 51, 73, 49, 31, 69, 74, 72, 59, 72, 81, 99, 101, 73

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{1067}{15} = 71.13$

Sum of squares of differences = 4739.733

$S.D = \sqrt{\frac{4739.733}{14}} = 18.39$

Population mean, μ = 60 minutes

Sample mean, $\bar{x}$ = 71.13 minutes

Sample size, n = 15

Alpha, α = 0.10

Sample standard deviation, s = 18.39 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 60\text{ minutes}\\H_A: \mu \neq 60\text{ minutes}$

We use Two-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{71.13 - 60}{\frac{18.39}{\sqrt{15}} } = 2.34$

Calculating the p-value from the table, we have,

P-value = 0.034354

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, we conclude that it does not appear that, as a group, the students are reasonably good at estimating one minute.

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3 years ago

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Answer:

$volume = \pi {r}^{2} h \\ = \pi( {5}^{2} ) \times 6 \\ 150\pi \: cubic \: units$

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3 years ago

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