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telo118 [61]
3 years ago
14

8. Anthony and Yareli's combined driving distance this week was 60 miles. Yareli drove 4 times as far as Anthony. How

Mathematics
2 answers:
NeX [460]3 years ago
8 0
You divide 60 and 4. And you get the answer
OLga [1]3 years ago
5 0

Answer:  Anthony drove 12 miles and Yareli drove 48 miles.

Step-by-step explanation:

let a = the distance Anthony drove

let y = the amount of time yareli drove

a + y = 60

y = 4a

since 4a is equal to y, you can substitute it in for the y’s in the first equation

a + 4a = 60

5a = 60

5a/5 = 60/5

a = 12

so Anthony drove 12 miles

now let’s plug that result into the y=4a equation to find out how many miles Yareli drove

y = 4a

y = 4(12)

y = 48

now let’s check to see if it works

48 + 12 = 60

12 x 4 = 48

it works :)

so Anthony drove 12 miles and Yareli drove 48 miles.

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Two numbers have theses properties . both numbers are greater than 6 . their hcf = 6 . their lcm = 60. what are the two numbers
aleksandrvk [35]
Let the numbers be x and y.
x*y=HCF*LCM=6*60=360
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Answer:

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

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 = 6, \sigma = 0.2

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

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

Z = \frac{6-6}{0.2}

Z = 0

Z = 0 has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

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

Z = \frac{6.2-6}{0.2}

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Z = 1 has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X  5.8.

X = 6.2

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

Z = \frac{6.2-6}{0.2}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 5.8

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

Z = \frac{5.8-6}{0.2}

Z = -1

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This is 1 subtracted by the pvalue of Z when X = 5.7.

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This is 1 subtracted by the pvalue of Z when X = 5.

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

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