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3 years ago

Having trouble with this word problem;

Mathematics

2 answers:

NeX [460]3 years ago

8 0

Answer:

If Mary runs at 5mhp and Janet runs at 7mph, it will take them 45 minutes to reach a point where there is 1.5 miles between them.

Or, if you'd like a shorter answer:

45 minutes.

Step-by-step explanation:

It will take both of them 1 hour to reach 2 miles between them. We are looking for 1.5 miles between them. So, split 1 hour into 4 equal parts.

1hr = 60 min.

4 x 15 = 60

Next, Add 3 of those 4 parts together or multiply one part by 3.

15 x 3 = 45

There's the answer.

Gwar [14]3 years ago

7 0

Answer:

8mph

Step-by-step explanation:

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-3x -10=32 can anybody help me

pochemuha

Answer:

x=-14

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

A 9-yard roll of string costs $3.42. What is the unit price?

Karolina [17]

Answer:

$0.38 per yard

Step-by-step explanation:

Step 1:

Given; A 9-yard roll of string that costs $3.42.

Required; To find the unit price

Step 2:

Convert $3.42 to cents

$1 ~dollar=100~cents ~\\therefore,\\$3.42=3.42~x~100=342~cents~$

Step 3:

Find the unit price

$\frac{9 yard}{1 yard} =\frac{342 ~cents}{x~cents}$

$9x=342$

$\frac{9x}{9} =\frac{342}{9}$

$x=38~cents$

Convert to dollars

$x=\frac{38}{100}$

$x=0.38~per~yard$

Hence the unit price is $0.38 per yard.

5 0

2 years ago

A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true prop

attashe74 [19]

Answer:

$0.424 - 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.286$

$0.424 + 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.562$

The 99% confidence interval would be given by (0.286;0.562)

Step-by-step explanation:

Information given:

$X= 36$ represent the families owned at least one DVD player

$n= 85$ represent the total number of families

$\hat p=\frac{36}{85}= 0.424$ represent the estimated proportion of families owned at least one DVD player

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.05$ . And the critical value would be given by: