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2 years ago

15

if you throw a rubber ball toward the ground at an angle, it will bounce up feom the ground at the same angle. find the unknown

angle measure in the drawing at the right.

1 answer:

Sauron [17]2 years ago

5 0

30°

A strait line is 180°
75°+75°=150°
180°-150°=30°

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In a recent year, there was about 400,000,000 mobile internet users. ( look at the picture )

AVprozaik [17]

$\bf 400,000,000\implies 4\times 10^8 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{\textit{desktop users}}{\textit{mobile users}}\qquad \qquad \cfrac{1.2\times 10^9}{4\times 10^8}\implies \cfrac{12\times 10^8}{4\times 10^8}\implies \cfrac{12}{4}\times\cfrac{10^8}{10^8}\implies \cfrac{3}{1}$

3 : 1, or 3 to 1, thus 3 times as many.

7 0

2 years ago

Read 2 more answers

I Need Help thanks <br>:)?

Feliz [49]

Answer:

6) x = 40.5

8) x = 16.5

Step-by-step explanation:

6) 45/x = 20/18

simplify and cross-multiply to get:

45/x = 10/9

10x = 405

x = 40.5

8) 22/8 = x/6

11/4 = x/6

4x = 66

x = 33/2

x = 16.5

8 0

2 years ago

If f(x) and g(x) are polynomial functions, what is the domain of f(x) + g(x) ?

IRINA_888 [86]

Answer:

all real numbers or $(-\infty, \infty)$

Step-by-step explanation:

since both f(x) and g(x) are polynomials, neither of them have restrictions on their domains. Since they can be defined for all real numbers. It's not like the square root which has a domain of (0, infinity), or log x which is defined only from (0, infinity). So adding these two polynomials would result in a domain of all real numbers.

7 0

1 year ago

Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 11, \sigma = 3$

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 25, s = \frac{3}{\sqrt{25}} = 0.6$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.6}$

$Z = 0.33$

$Z = 0.33$ has a pvalue of 0.6293.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.6}$

$Z = -0.33$

$Z = -0.33$ has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

$Z = \frac{11 - 11}{0.6}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 10.5

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

$Z = \frac{10.5 - 11}{0.6}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 100, s = \frac{3}{\sqrt{100}} = 0.3$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.3}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0

2 years ago

The distance to the nearest exit door is less than 150 feet.

Alja [10]

Answer:

d<150

Step-by-step explanation:

6 0

2 years ago