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3 years ago

Shareen walked a total of 8.96 miles in a walk-a-thon. If her average speed was 2.8 miles per hour, how

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

3 0

Answer:

3 hours and 12 minutes

Step-by-step explanation:

8.96 ÷ 2.8= 3.2

I know that 2/10 of 60 is 12

3.2= 3hrs & 12 mins

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(Picture) MULTIPLYING POLYNOMIALS AND SIMPLIFYING EXPRESSIONS PLEASE HELP!!!!

creativ13 [48]

Answer:

$12a^3 +2a^2 +23a+18$

Step-by-step explanation:

$(3a+2)(4a^2-2a+9)$

we multiply 3a+2 inside the parenthesis to find the product

first multiply 3a inside second parenthesis

$12a^3 - 6a^2 +27a$

now multiply +2 inside second parenthesis

$8a^2 - 4a + 18$

Now we combine it

$12a^3 - 6a^2 +27a+8a^2-4a+18$

combine like terms

$12a^3 +2a^2 +23a+18$

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=3%20%5Ctimes%204x%20%5Ctimes%202y" id="TexFormula1" title="3 \times 4x \times 2y" alt="3 \time

Dahasolnce [82]

Answer:

24 x y

Step-by-step explanation:

Simplify the following:

3×4×2 x y

Hint: | Multiply 3 and 4 together.

3×4 = 12:

12×2 x y

Hint: | Multiply 12 and 2 together.

12×2 = 24:

Answer: 24 x y

5 0

3 years ago

Phone plan A charges $1.25 for the first minute and $0.15 for every minute thereafter. Phone plan B charges a $0.90 connection f

melomori [17]

Answer:

The answer is (C) $1.25+0.15(x-1)=0.90+0.20x$ .

Step-by-step explanation:

Now, to find the equation for finding the length, in minutes, of a phone call that costs the same under either plan.

Let the phone charges per minute be $x.$

So, to get the equation of the phone plan A as the phone plan A charges $1.25 for the first minute and after that $0.15 for every minute:

$1.25+0.15x-0.15$

$=1.25-0.15(x-1)$

Now, to get the equation of the phone plan B as given phone plan B charges a $0.90 connection fee and $0.20 per minute:

$0.90+0.20x$

As, the phone call that costs the same under either plan.

Thus:

$1.25+0.15(x-1)=0.90+0.20x$

Therefore, the answer is (C) $1.25+0.15(x-1)=0.90+0.20x$ .

4 0

4 years ago

Four women must cross a bridge over a deep ravine in enemy territory in the middle of the night, treacherous bridge will hold on

Galina-37 [17]

Ok, this is usually a trial and error question and it is supposed to be fun.
I believe you want the four women to cross the bridge in the least possible time before the bridge collapses.

The solution is as follows:
1- The 5-minute lady and 10-minute lady cross the bridge
(the total time will be 10 minutes)
2- The 5-minute lady returns
(the total time will be 5 + 10 = 15 minutes)
3- The 20-minute lady and 25-minute lady cross
(the total time will be 15 + 25 = 40 minutes)
4- The 10-minute lady returns
(the total time will be 10 + 40 = 50 minutes)
5- The 5-minute lady and 10-minute lady finally cross
(the total time will be 10 + 50 = 60 minutes)

The last one to cross should step aside from the bridge as quickly as possible before the bridge collapses.

8 0

4 years ago

Consider a population proportion p = 0.68. Calculate the expected value and the standard error of with n = 20. Is it appropriate

zubka84 [21]

Answer: a) 0.68, 0.1043, b) 0.68, 0.066

Step-by-step explanation:

Since we have given that

p = 0.68

n= 20

Expected value = 0.68

So, Standard error would be

$\sqrt{\dfrac{p(1-p)}{n}}\\\\=\sqrt{\dfrac{0.68\times 0.32}{20}}\\\\=0.1043$

if n = 50

Expected value = 0.68

So, Standard error would be

$\sqrt{\dfrac{p(1-p)}{n}}\\\\=\sqrt{\dfrac{0.68\times 0.32}{50}}\\\\=0.0659\approx 0.066$

In both the cases,

Yes it is appropriate to use the normal distribution for expected value and standard error.

5 0

4 years ago