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4 years ago

the track coach tells Dave that he has 84% of his workout left. Dave has run 4 laps so far. How many laps will Dave run in all

1 answer:

4 0

If 4 is 100-84=16 % of his workout, 4 is 16/100 or 4/25 of his workout. If 4=4x/25, where x is his workout length, dividing by 4 and multiplying by 25 gives x=25, so Dave must run 25 laps total.

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Plz help me i need to pass I've been trying this lesson all day and I can't figure this out

Tasya [4]

Step-by-step explanation:

step 1: given= 6*0=0

step 2: Use the distributive property

step 3: Substitute

step 4: -

step 5: Subtract 60 from both sides

step 5: simplify

7 0

3 years ago

Read 2 more answers

Assume the average cost of maintaining a dog is the same from month to month. After 55 months, the dog had cost you a total of $

Marizza181 [45]

Answer:

Originally, you will <em>pay</em> $95.75 for the dog.

Step-by-step explanation:

Given:

Maintenance cost of dog after 55 months = $5,468.50

Maintenance cost of dog after 82 months = $8,053.75

To Find:

How much did you pay for the dog originally = ?

Solution:

It is stated that the maintenance cost is same from month to month

Now 82 - 55 = 27 months

For 27 we have paid

$8,053.75 - $5,468.50

$2585.25

But one month the amount paid will be

$\frac{2585.25}{27}$

7 0

3 years ago

How to write slope-intercept form through (-1,3) and a slope of -1

oksian1 [2.3K]

The slope-intercept form:

$y=mx+b$

m - slope

b - y-intercept

We have the slope m = -1 and the point (-1, 3). Substitute:

$3=(-1)(-1)+b$

$3=1+b$ <em>subtract 1 from both sides</em>

$2=b\to b=2$

<h3>Answer: y = -x + 2</h3>

7 0

4 years ago

Does the function y = 8x represent a direct or inverse variation

liraira [26]

This is direct variation since the x isn't in the denominator.

3 0

3 years ago

A bag contain 3 black balls and 2 white balls.

Troyanec [42]

Answer:

Step-by-step explanation:

Total number of balls = 3 + 2 = 5

$Probability \ of \ taking \ 2 \ black \ ball \ with \ replacement\\\\ = \frac{3C_1}{5C_1} \times \frac{3C_1}{5C_1} =\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\\\\$

$Probability \ of \ one \ black \ and \ one\ white \ with \ replacement \\\\= \frac{3C_1}{5C_1} \times \frac{2C_1}{5C_1} = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$

Probability of at least one black( means BB or BW or WB)

$=\frac{3}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5} \\\\= \frac{9}{25} + \frac{6}{25} + \frac{6}{25}\\\\= \frac{21}{25}$

Probability of at most one black ( means WW or WB or BW)

$=\frac{2}{5} \times \frac{2}{5} + \frac{3}{5} \times \frac{2}{5} \times \frac{2}{5} + \frac{3}{5}\\\\= \frac{4}{25} + \frac{6}{25} + \frac{6}{25}\\\\=\frac{16}{25}$

a) Probability both black without replacement

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

b) Probability of one black and one white

$=\frac{3}{5} \times \frac{2}{4}\\\\=\frac{6}{20}\\\\=\frac{3}{10}$

c) Probability of at least one black ( BB or BW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{6}{20} + \frac{6}{20} \\\\=\frac{18}{20} \\\\=\frac{9}{10}$

d) Probability of at most one black ( BW or WW or WB)

$=\frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{1}{4} + \frac{2}{5} \times \frac{3}{4}\\\\=\frac{6}{20} + \frac{2}{20} + \frac{6}{20} \\\\=\frac{14}{20}\\\\=\frac{7}{10}$

6 0

3 years ago