Two angles are vertical angles. One is labeled 2x. The other angle is labeled (x+30). Find the value of x.

What principal will amount to $3000 if invested at 3% interest compounded

MakcuM [25]

Answer: $2227.41

Step-by-step explanation:

A = P(1 + r/n)^nt

4 0

3 years ago

A researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playi

Andreas93 [3]

Answer:

Proportion of the students recalled more than 15 names is 91.77%.

Step-by-step explanation:

We are given that a researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playing the name game, the students were asked to recall as many first names of fellow students as possible.

The mean number of names recalled was 19.41 with a standard deviation of 3.17.

Let X = number of names recalled

SO, X ~ N( $\mu = 19.41,\sigma^{2} = 3.17^{2}$ )

The z-score probability distribution is given by ;

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

where, $\mu$ = mean number of names recalled = 19.41

$\sigma$ = standard deviation = 3.17

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.

Now, proportion of the students recalled more than 15 names is given by = P(X > 15 names)

P(X > 15) = P( $\frac{X-\mu}{{\sigma} } }$ > $\frac{15-19.41}{3.17} }$ ) = P(Z > -1.39)

= P(Z < 1.39) = 0.9177 {using z table}

Therefore, proportion of the students recalled more than 15 names is 91.77%.

4 0

3 years ago

Cheddar cheese cost $4.25 per pound. Which equation best represents y, the total cost of x pounds of cheddar cheese? Group of an

user100 [1]

Answer:

I'm going to edit this later, but do you have the full question?

6 0

3 years ago

Iq scores are normally distributed with a mean of 100 and a standard deviation of 15.

bearhunter [10]

Mean is $\mu=100$ and a standard deviation is $\sigma =15$ , then the variable $X\sim N(100,15^2)$ .

Use substitution

$Z=\dfrac{X-\mu}{\sigma},\\ \\ Z=\dfrac{X-100}{15}$ .

This substitution gives you that $Z\sim N(0,1)$ .

a. For X=130, $Z=\dfrac{130-100}{15}=2$ and $Pr(X>130)=Pr(Z>2) =0.9772$ (the decimal value is taken from the Standard Normal Distribution Table).

b. For X=90, $Z=\dfrac{90-100}{15}=-\dfrac{2}{3}$ and for X=110, $Z=\dfrac{110-100}{15}=\dfrac{2}{3}$ . Then $Pr(90$ (the decimal value is taken from the Standard Normal Distribution Table).