All categories

2 years ago

Ariel recorded the number of e-mails she received

Mathematics

1 answer:

gizmo_the_mogwai [7]2 years ago

4 0

D. 24
Mode is the number most often repeated in the sequence.

You might be interested in

Youssef plans to get rid of cable and is comparing two high-speed internet providers. He narrows his search to two companies. Co

nadezda [96]

100 per setup fee. a.

4 0

3 years ago

Find the explicit formula for the given sequence 1/4, 3/4, 5/4, 7/4

Harlamova29_29 [7]

Step-by-step explanation:

d=(3/4)-(1/4)=(5/4)-(3/4)=(7/4)-(5/4)=1/2

then it is an arithmetic sequence

by using this equation :

a(n) = a1 +(n-1)d

and we have : a1=(1/4), d=(1/2)

So:

Explicit formula: a(n) = (1/4) +(n-1)*(1/2)

<em>hope this helps </em>

<em />

8 0

2 years ago

Read 2 more answers

Find the slope of a line that passes through the given points. Write your answer below.

12345 [234]

Just use the formula:
$slope(m) = \frac{y_2-y_1}{x_2-x_1}$

Then, plug in numbers:
$m = \frac{1-4}{-2-3} = \frac{-3}{-5} = \frac{3}{5}$

And that would be your answer: 3/5 is the slope.

4 0

3 years ago

Delaney would like to make a 5 lb nut mixture that is 60% peanuts and 40% almonds. She has several pounds of peanuts and several

Alchen [17]

Answer:

a) The system that models the situation is:

p + 0.2m = 3

p + m = 5

b) The solution of the system is:

m = 2.5

p = 2.5

Step-by-step explanat:

It is desired to obtain 5 pounds of a mixture containing 40% peanuts and 60% almonds.

It has a mixture of 20% peanuts and 80% almonds.

p: number of pounds of peanuts

m = amount of mixture

The equation that represents the amount of peanut and mixture that is needed is:

p + m = 5 pounds

m = 5-p (i)

The amount of peanut in total that the mixture should have is:

p + 0.2m = 0.6 (5)

p + 0.2m = 3 (ii)

a) The system that models the situation is:

p + 0.2m = 3

p + m = 5

Now we substitute (i) in (ii) and clear p.

p + 0.2 (5-p) = 3

p-0.2p = 2

p = 2.5 pounds

Now we find m

m = 5 - 2.5

m = 2.5 pounds

b) The solution of the system is:

m = 2.5

p = 2.5

7 0

2 years ago

A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is .5

Elodia [21]

Answer:

7.3% percentage of the bearings produced will not be acceptable.

Step-by-step explanation:

Consider the provided information.

Average diameter of the bearings it produces is .500 inches. A bearing is acceptable if its diameter is within .004 inches of this target value.

Let X is the normal random variable which represents the diameter of bearing.

Thus, 0.500-0.004<X<0.500+0.004

0.496<X<0.504

The bearings have normally distributed diameters with mean value .499 inches and standard deviation .002 inches.

Use the Z score formula: $\frac{X-\mu}{\sigma}$

Therefore

$\frac{0.496-0.499}{0.002}\leq z\leq \frac{0.504-0.499}{0.002}$

$\frac{-0.003}{0.002}\leq z\leq \frac{0.005}{0.002}$

$-1.5\leq z\leq 2.5$

Now use the standard normal table and determine the probability of that a ball bearing will be acceptable.

$P(-1.5\leq z\leq 2.5)=0.9938-0.0668=0.9270$

We need to find the percentage of the bearings produced will not be acceptable.

So subtract it from 1 as shown.

1-0.9270=0.073

Hence, 7.3% percentage of the bearings produced will not be acceptable.

3 0

3 years ago