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2 years ago

Xavier rides his bike 5 1/4 miles or 82% of his goal for the day. What is his goal for the day?

Mathematics

1 answer:

Leno4ka [110]2 years ago

8 0

Use ratio and proportion

5.25/x= 82/100
cross multiply
82 x= 525
x= 6.4 miles

Hope this helps

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The weight of 3 eggs is shown. Identify the constant of proportionally of total weight to number of eggs. The weight of the eggs

HACTEHA [7]

It would be 48g per egg

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2 years ago

What facts are required in order to conclude that m∠ADE = m∠ABC?

mariarad [96]

Answer:

The correct option is;

DE = 2·(BC), AD = 2·(AB), and AE = 2·(AC)

Step-by-step explanation:

Given that we have;

1) The side AD of the angle m∠ADE corresponds to the side AB of the angle m∠ABC

2) The side DE of the angle m∠ADE corresponds to the side BC of the angle m∠ABC

3) The side AE of the angle m∠ADE corresponds to the side AC of the angle m∠ABC

Then when we have DE = 2·(BC), AD = 2·(AB), and AE = 2·(AC), we have by sin rule;

AE/(sin(m∠ADE)) = 2·(AC)/(sin(m∠ABC)) = AE/(sin(m∠ABC))

∴ (sin(m∠ADE)) = (sin(m∠ABC))

m∠ADE) = m∠ABC).

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2 years ago

Chapter 2

Mashutka [201]

Answer:

11.20

Step-by-step explanation:

Multiply both equations by 100:

19.99=1999

8.79=879

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Divide by 100:

1120/100=11.20 saved ................

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2 years ago

Translate this sentence into an equation.

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2 years ago

Read 2 more answers

a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was

Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

Let $\bar X$ = sample average time spent studying

The z-score probability distribution for sample mean is given by;

Z = $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean hours spent studying = 25 hours

$\sigma$ = standard deviation = 15 hours

n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < $\bar X$ < 30 hours)

P(29 hours < $\bar X$ < 30 hours) = P( $\bar X$ < 30 hours) - P( $\bar X$ $\leq$ 29 hours)

P( $\bar X$ < 30 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ < $\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z < 2) = 0.97725

P( $\bar X$ $\leq$ 29 hours) = P( $\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }$ ) = P(Z $\leq$ 1.60) = 0.94520

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.

Therefore, P(29 hours < $\bar X$ < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0

2 years ago