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2 years ago

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A plane flew for 4 hours heading south and for 6 hours heading east. If the total distance traveled was 3,370 miles, and the pla

ne traveled 45 miles per hour faster heading south, at what speed was the plane traveling east?

1 answer:

matrenka [14]2 years ago

7 0

We know the total distance and individual times, so we can sum and equate total distance.

Let
E=speed heading east
E+45=speed heading south
then
4(E+45)+6E=3370 miles
Solve for E:
4E+180+6E=3370
10E=3370-180=3190 mph
E=319 mph.

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What value of n makes the equation true? -1/5n + 7 =2

klasskru [66]

The answer is 25.
-1/5n = -5
-n = -25
n = 25

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1 year ago

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Triangle S R Q is shown. Angle S R Q is a right angle. An altitude is drawn from point R to point T on side S Q to form a right

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Answer:

Option B.

Step-by-step explanation:

It is given that ΔSRQ is a right angle triangle, ∠SRQ is right angle.

RT is altitude on side SQ, ST=9, TQ=16 and SR=x.

In ΔSRQ and ΔSTR,

$m\angle S=m\angle S$ (Reflexive property)

$m\angle R=m\angle T$ (Right angle)

By AA property of similarity,

$\triangle SRQ\sim \triangle STR$

Corresponding parts of similar triangles are proportional.

$\dfrac{SR}{SQ}=\dfrac{ST}{SR}$

Substitute the given values.

$\dfrac{x}{9+16}=\dfrac{9}{x}$

$\dfrac{x}{25}=\dfrac{9}{x}$

On cross multiplication we get

$x^2=25\times 9$

$x^2=225$

Taking square root on both sides.

$x=\sqrt{225}$

$x=15$

The value of x is 15. Therefore, the correct option is B.

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2 years ago

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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.

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2 years ago

Steve stayed after school for extra activities 55% of the 180 school days this year. How many days did Steve stay after school?

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Answer:

11/20 or 99/180 days

Step-by-step explanation:

55/100 converted to ?/180

180/100=1.8

so you would do the same on the numerator which is 55*1.8=99

99/180= 33/60=11/20

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1 year ago

Find the residual values, and use the graphing calculator tool to make a residual plot. A 4-column table with 5 rows. The first

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Answer:

Solution-

We know that,

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The table for residual values is shown below,

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We know that,

If the points in a residual plot are randomly dispersed around the horizontal or x-axis, a linear regression model is appropriate for the data. Otherwise, a non-linear model is more appropriate.

As, in this case the points are distributed randomly around x-axis, so the residual plot show that the line of regression is best fit for the data set.

Hope this helps!

Step-by-step explanation:

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2 years ago

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