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2 years ago

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Maya chatted online with all her friends for 2/3 of an hour on Saturday and 1/4 of an hour on sunday. During that time period, s

he chatted with paul for 1/8 of an hour. what fraction of the time maya spent chatting with her friends online was spent chatting with paul?

2 answers:

Kamila [148]2 years ago

8 0

Time spent with paul : 1/8 h
Total time spent chatting with friends: 2/3h+1/4h=11/12 h

The fraction of the time she spent with paul is (time spent with paul)/(total time spent chatting)

hence the fraction spent with paul is (1/8)/(11/12)=3/11

lisabon 2012 [21]2 years ago

4 0

Answer:

2/3 = 16/24

1/4 = 6/24

1/8 = 3/24

(3/24) / (25/24)

(3/24) * (24/25) = 72/600 = 12/100 = 3/25

100 * (1/8)/ (8/12+3/12)

= 100 * (1/8) / (11/12)

= 100 * (3/24)(22/24)

= 100 (3/22)

= 13.6%

Step-by-step explanation:

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Alex_Xolod [135]

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2 years ago

Lynn is making custom bricks. She combines 39 pounds of water and 48 pounds of brick dust in a mixing bucket. She spills 13 poun

ehidna [41]

Lynn mixes 39 pounds of water with 48 pounds of brick dust, so the mixture has a total weight of 39 + 48 = 87 pounds.

13 pounds are spilled during mixing, so she's left with 87 - 13 = 74 pounds to turn into bricks.

7 pounds are needed for 1 brick, so she can make 10 bricks because 7 • 10 = 70 and 7 • 11 = 77, and 70 < 74 < 77.

If she uses the mixture to make 10 bricks, she uses up 70 pounds of it, which leaves 4 pounds to be washed out.

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2 years ago

1) The roots of the equation 9x2 + 3x – 4 = 0 are

Wittaler [7]

A !! hope this helps

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2 years ago

Read 2 more answers

The XO Group Inc. conducted a survey of 13,000 brides and grooms married in the United States and found that the average cost of

slavikrds [6]

Answer:

a) 0.0392

b) 0.4688

c) At least $39,070 to be among the 5% most expensive.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 29858, \sigma = 5600$

a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?

This is the pvalue of Z when X = 20000. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20000 - 29858}{5600}$

$Z = -1.76$

$Z = -1.76$ has a pvalue of 0.0392.

So this probability is 0.0392.

b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?

This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.

X = 30000

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30000 - 29858}{5600}$

$Z = 0.02$

$Z = 0.02$ has a pvalue of 0.5080.

X = 20000

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20000 - 29858}{5600}$

$Z = -1.76$

$Z = -1.76$ has a pvalue of 0.0392.

So this probability is 0.5080 - 0.0392 = 0.4688

c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?

This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 29858}{5600}$

$X - 29858 = 5600*1.645$

$X = 39070$

The wedding would have to cost at least $39,070 to be among the 5% most expensive.

5 0

2 years ago

the perimeter of a rectangular campsite is 64m and its area is 207m^2. find the length and breadth of the campsite

NNADVOKAT [17]

Answer:

The length is $23\ m$ and the breadth is $9\ m$

Step-by-step explanation:

Let

x ----> the length of a rectangular campsite

y ----> the breadth of a rectangular campsite

we know that

The perimeter of a rectangle is equal to

$P=2(x+y)$

$P=64\ m$

so

$64=2(x+y)$

$32=(x+y)$

$y=32-x$ ------> equation A

The area of a rectangle is equal to

$A=xy$

$A=207\ m^{2}$

so

$207=xy$ -----> equation B

substitute equation A in equation B

$207=x(32-x)\\\\207=32x-x^{2}\\\\x^{2}-32x+207=0$

Solve the quadratic equation by graphing

The solution is $x=23\ m$ ( I assume that the length is greater than the breadth)

see the attached figure

Find the value of y

$y=32-23=9\ m$

therefore

The length is $23\ m$ and the breadth is $9\ m$

4 0

2 years ago