Write the equation of the line in slope-intercept form that has the following slope and intercept: `m=-\frac{1}{2}` and `b=-2`

Erik buys 2.5 pounds of cashews.If each pound of cashews $7.70,how much will he pay for the cashews???I put Erik will pay $49250

Fittoniya [83]

The answer should be $ 19.25. All you have to do is $7.70 * 2.5 which = $19.25

3 0

2 years ago

Read 2 more answers

What is the radius of a circle if the area is 78.5 cm^2

Verizon [17]

78.5= (3.14)r^2
25=r^2
5=r
So, the radius of a circle with an area is 78.5 cm^2 is 5 cm^2

3 0

2 years ago

Suppose you have the following recursion formula a1 = 1, a2 = 2, and an = a(n - 1)+ a(n - 2)for integers n ≥ 3. How would you de

julia-pushkina [17]

A3=3
a4=5
a5=8

This sequence has properties similar to those of the Fibonacci sequence.

8 0

2 years ago

A researcher determines that students are active about 60 + 12 (M + SD) minutes per day. Assuming these data are normally distri

rjkz [21]

Answer:

The correct option is (b).

Step-by-step explanation:

If X $\sim$ N (µ, σ²), then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z $\sim$ N (0, 1).

The distribution of these z-variate is known as the standard normal distribution.

The mean and standard deviation of the active minutes of students is:

μ = 60 minutes

σ = 12 minutes

Compute the z-score for the student being active 48 minutes as follows:

$Z=\frac{X-\mu}{\sigma}=\frac{48-60}{12}=\frac{-12}{12}=-1.0$

Thus, the z-score for the student being active 48 minutes is -1.0.

The correct option is (b).

4 0

2 years ago

Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that

amm1812

Answer:

The value is $P(| \^ p - p| < 0.05 ) = 0.9822$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.52$

The sample size is n = 563

Generally the population mean of the sampling distribution is mathematically represented as

$\mu_{x} = p = 0.52$

Generally the standard deviation of the sampling distribution is mathematically evaluated as

$\sigma = \sqrt{\frac{ p(1- p)}{n} }$

=> $\sigma = \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }$

=> $\sigma = 0.02106$

Generally the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

$P(| \^ p - p| < 0.05 ) = P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 ))$

Here $\^ p$ is the sample proportion of persons with a college degree.

So

$P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma } < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )$