Evaluate the expression using the order of operation (13.1 times 3) + 5.21

A number divided by three list two is at most two .

d1i1m1o1n [39]

Is this the original question or?

5 0

3 years ago

Johnathan bought a new computer for $1,716 using the electronics store’s finance plan. He will pay $143 a month for 12 months. W

DENIUS [597]

Answer:

1,1716 + 143(12)

Step-by-step explanation:

i believe that is correct hope I helped you!

4 0

2 years ago

= 1/2 (u+v)t

RoseWind [281]

The distance travelled by an object is 50 m.

<h3>What is Distance?</h3>

The distance between two points is the length of the path connecting them.

Here, given equation of distance

S = 1/2 (u + v)t

Where s is the distance traveled by an object (in meters)

initial velocity u (in m/s)

final velocity v (in m/s).

t in seconds.

Now, given values are;

u = 8 m/s ; v = 12 m/s ; t = 5 sec.

S = 1/2 (8 + 12)5

S = 1/2 (20)5

S = 100/2

S = 50 m.

Thus, the distance travelled by an object is 50 m.

Learn more about Distance from:

brainly.com/question/15172156

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5 0

2 years ago

Suppose a 90% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$38,737, $50,46

JulsSmile [24]

Answer:

a. The point estimate was of $44,600.

b. The sample size was of 16.

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

The margin of error is the difference between the two bounds, divided by 2.

a. What is the point estimate of the mean salary for all college graduates in this town?

Mean of the bounds, so:

(38737+50463)/2 = 44600.

The point estimate was of $44,600.

b. Determine the sample size used for the analysis.

First we need to find the margin of error, so:

$M = \frac{50463-38737}{2} = 5863$

Relating the margin of error with the sample size:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.05 = 0.95$ , so Z = 1.64.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

For this problem, we have that $\sigma = 14300, M = 5863$ . So

$M = z\frac{\sigma}{\sqrt{n}}$

$5863 = 1.645\frac{14300}{\sqrt{n}}$

$5863\sqrt{n} = 1.645*14300$

$\sqrt{n} = \frac{1.645*14300}{5863}$

$(\sqrt{n})^2 = (\frac{1.645*14300}{5863})^2$

$n = 16$

The sample size was of 16.

7 0

3 years ago

Decimals between 10 and 15 with an interval of .75 between each pair of decimals

d1i1m1o1n [39]

This is a sum that needs understanding the starting and the ending number first. then you should understand that between the starting and the ending number there has to be pairs of numbers with a difference of 0.75.
The numbers are:
10.75, 11.50, 12.25, 13.00, 13.75, 14.50.

6 0

3 years ago