In the past year, Hans watched 38 movies that he thought were very good. He watched 95 movies over the whole year. Of

Jina wants to measure the width of a river. She marks off two right triangles, as shown in the figure. The base of the larger tr

Brilliant_brown [7]

Answer:

width of a river = 45m

Step-by-step explanation:

ration and proportion

let x = width of a river

x 20.9 m

------ = --------

56 m 26 m

x = (20.9 * 56) / 26

x = 45 m

therefore the width of a river is 45 m

7 0

3 years ago

Sin150+sin120÷cos210-cos300

natulia [17]

From sin(180-x) = sin(x)
sin(150) = sin(180-30) = sin(30)
sin(120) = sin(180-60) = sin(60)

from cos(360-x) = cos(x)
cos(300) = cos(360-60) = cos(60)
cos(210) = cos(360-150) = cos(150)

from cos(180-x) = -cos(x)
cos(210) = cos(150) = cos(180-30) = -cos(30)

6 0

3 years ago

Please help me out!

seropon [69]

Answer:

Second option

Third option

Fourth option

Step-by-step explanation:

We have the following quadratic function

$f(x) =(x-1)(x+7)$

Use the distributive property to multiply the expression

$f(x) =x^2+7x-x-7$

$f(x) =x^2+6x-7$

For a function of the form $f (x) = ax ^ 2 + bx + c$ the x coordinate of the vertex is:

$x =-\frac{b}{2a}$

Then in this case the coordinate of the vertex is:

$x =-\frac{6}{2(1)}$

$x =-3$

To obtain the y coordinate of the vertex we evaluate the function at $x = -3$

$f(-3) =(-3)^2+6(-3)-7$

$f(-3) =9-18-7$

$f(-3) =-16$

Then the vertex is: (-3, -16)

We can see in the graph that the zeros of the function are x=1 and x=-7

Then the function is decreasing from -∞ to -3 and then it is increasing from -3 to ∞

The function is positive for $x$ and $x> 1$

The correct answers are:

Second option

Third option

Fourth option

7 0

2 years ago

The first term in the sequence is 1, and each other term is 5 more than three times the previous term.

marin [14]

$a_1=1, a_n=a_{n-1}+5$ ( $n>1$ ).

Hope this helps.

6 0

2 years ago

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0

3 years ago