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Alex
3 years ago
12

How many miles can you get on one tank of gas if it holds 18 gallons and you get 22 miles per gallon

Mathematics
1 answer:
Alexeev081 [22]3 years ago
4 0
396

18 x 22= 396 hope i helped you.
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2.87 pound sterling
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3 years ago
Easy question!
forsale [732]

Answer:

  • yes
  • 5 : 2 or 2.5 : 1

Step-by-step explanation:

The ratios of corresponding linear dimensions are identical:

  5/2 = 15/6 = 2.5

so we can conclude the figures are similar. The scale factor is the ratio of corresponding dimensions:

  first : second = 5 : 2 . . . or . . . 2.5 : 1

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3 years ago
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HELP MEE!!!! Yesterday, the temperature at noon was 11.4℉. By midnight, the temperature had decreased by 15.7 degrees. What was
Kruka [31]

Answer:

The answer is the first one, -4.3 degrees

Step-by-step explanation:

Please give me brainliest!

7 0
3 years ago
The manager of a new supermarket wished to estimate the likely expenditure of his customers. A sample of till slips from a simil
bagirrra123 [75]

Answer:

0.0228 = 2.28% probability that any shopper selected at random spends more than $80 per week.

88.54% of shoppers are expected to spend between $30 and 80 per week.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean of $50 and a standard deviation of $15.

This means that \mu = 50, \sigma = 15

Find the probability that any shopper selected at random spends more than $80 per week?

This is 1 subtracted by the p-value of Z when X = 80. So

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

Z = \frac{80 - 50}{15}

Z = 2

Z = 2 has a p-value of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that any shopper selected at random spends more than $80 per week.

Find the percentage of shoppers who are expected to spend between $30 and 80 per week?

The proportion is the p-value of Z when X = 80 subtracted by the p-value of Z when X = 30.

X = 80

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

Z = \frac{80 - 50}{15}

Z = 2

Z = 2 has a p-value of 0.9772

X = 30

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

Z = \frac{30 - 50}{15}

Z = -1.33

Z = -1.33 has a p-value of 0.0918

0.9772 - 0.0918 = 0.8854

0.8854*100% = 88.54%

88.54% of shoppers are expected to spend between $30 and 80 per week.

8 0
3 years ago
95i is a root of f(x)=x^2 +9025. find the other roots of f(x)
Alex_Xolod [135]

Answer:

Other root is -95i

Step-by-step explanation:

Here, Nature of roots of f(x) is imaginary roots.

Therefore, Roots are conjugate of each other.

Conjugate of x + yi is x - yi.

we get conjugate of 95i as -95i

Verification shows....

(x-95)(x+95i)=0

{x }^{2}  -  {95}^{2}  {i}^{2}  = 0 \\  {x }^{2}   + {95}^{2} = 0  \\ {x }^{2}   + 9025= 0   \\ f(x) = 0

this, roots are -95i and 95i

5 0
4 years ago
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