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sdas [7]
2 years ago
9

Elizabeth can run 3/4 mile in 1/10 of an hour. What is her speed in miles per hour?

Mathematics
1 answer:
Temka [501]2 years ago
7 0
0.75 miles per hour I think....
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Please help ASAP need answers thank you
wlad13 [49]

Similar triangles

x/3 = (9+3)/x

x/3 = 12/x

x^2 = 3 * 12

x^2 = 36

x = 6

Answer

6

6 0
2 years ago
I struggle in math, and occasionally I'll confuse the questions I try to answer
Alex17521 [72]
15X=12X+12
Subtract 12X
3X=12
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3 0
3 years ago
1 1/3 yards converted to inches
Semmy [17]
Google will do it for you. its 48 inches

4 0
3 years ago
Read 2 more answers
In p.
ozzi
The circumference by definition is given by:
 C = 2 * pi * r
 Where,
 r: radius of the circle.
 Substituting values we have:
 C = 2 * 3.14 * 16
 C = 100.48 feet
 Answer:
 
A measurement that is closest to the circumference of the parachute in feet is:
 
C = 100.48 feet
6 0
2 years ago
A videotape store has an average weekly gross of $1,158 with a standard deviation of $120. Let x be the store's gross during a r
statuscvo [17]

Answer:

The number of standard deviations from $1,158 to $1,360 is 1.68.

Step-by-step explanation:

Problems of normally distributed samples are 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 = 1158, \sigma = 120

The number of standard deviations from $1,158 to $1,360 is:

This is Z when X = 1360. So

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

Z = \frac{1360 - 1158}{120}

Z = 1.68

The number of standard deviations from $1,158 to $1,360 is 1.68.

3 0
2 years ago
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