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3 years ago

13

A rock is tossed over a seaside cliff. The equation y=−16x^2+50x+118 gives the height of a rock above sea level. How long until

the rock hits the water?

1 answer:

Karolina [17]3 years ago

8 0

I don’t know I just want to get points but good luck

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Last year 63 inches of rain fell oaktown. In pine city 7 inches of rain fell last year. Write and solve equation to find how man

Nonamiya [84]

Answer:

Oaktown had 9 times more inches of rain than Pine City did.

Step-by-step explanation:

63 divided by 7 = 9.

Very simple problem. Glad to help.

4 0

3 years ago

Suppose that the sitting back-to-knee length for a group of adults has a normal distribution with a mean of mu equals 24.4 in.

solong [7]

Answer:

$P(X \geq 26.6) = 0.0336$ , which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 24.4, \sigma = 1.2$

In this problem, a value x is significantly high if:

$P(X \geq x) = 0.01$

Using these criteria, is a back-to-knee length of 26.6 in. significantly high?

We have to find the probability of the length being 26.6 in or more, which is 1 subtracted by the pvalue of Z when X = 26.6. So

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

$Z = \frac{26.6 - 24.4}{1.2}$

$Z = 1.83$

$Z = 1.83$ has a pvalue of 0.9664.

1 - 0.9664 = 0.0336

$P(X \geq 26.6) = 0.0336$ , which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

6 0

3 years ago

Please answer this right now, beacause i really need it to submit this righttt now :(

erik [133]

1. 14:8, 21:12 28:16
2. 10/12 15/18 20/24
5. 5:4
6. No
7. No
8. Yes
9. No
10. 15

5 0

2 years ago

2 Addition of which two number is<br>736

Elina [12.6K]

Step-by-step explanation:

it can be 700 + 36

hope it helps...

7 0

3 years ago

Assume that the wavelengths of photosynthetically active radiations (PAR) are uniformly distributed at integer nanometers in the

Lana71 [14]

There are 26 integers in the set $\{630,631,\ldots,655\}$ . If the wavelength of light is uniformly distributed, then the probability that it has a wavelength of any one of these integers is $\dfrac1{26}$ , so the distribution of wavelengths has probability mass function

$\mathbb P(X=x)=\begin{cases}\dfrac1{26}&\text{for }630\le x\le655,x\in\mathbb Z\\\\0&\text{otherwise}\end{cases}$

The mean (expected value) is given by

$\mathbb E(X)=\displaystyle\sum_xx\mathbb P(X=x)$

and the variance by

$\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$

First, the mean:

$\mathbb E(X)=\displaystyle\sum_{x=630}^{655}x\mathbb P(X=x)=\frac1{26}\sum_{x=630}^{655}x$
$\mathbb E(X)=\dfrac{16705}{26}\approx643$

(rounded up from an exact value of 642.5)

Now for the variance:

$\mathbb E(X^2)=\displaystyle\sum_xx^2\mathbb P(X=x)=\frac1{26}\sum_{x=630}^{655}x^2$
$\mathbb E(X^2)=412862.5$

Meanwhile, $\mathbb E(X)^2=642.5^2=412806$ , which gives a total variance of

$\mathbb V(X)=412862.5-412806\approx57$

(rounded up from 56.5)

3 0

3 years ago