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kkurt [141]
4 years ago
14

What is the answer for this 90 = 6 (d - 84)

Mathematics
1 answer:
REY [17]4 years ago
4 0
Make sure D is by its self so
90=6(d-84)
Divide both sides by 6 to get rid of 6 from the right side so
90/6= (d-84)
15 =(d-84)
Now get rid of the 84 by adding it
15+84= d
99=d therefore d is equal to 99
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The units’ digit of a two-digit number is 5 more than the tens’ digit, and the number is three times as great as the sum of the
solniwko [45]

Answer:

The number is 27

Step-by-step explanation:

Let the 10s digit be x

Let the units digit be y

y = x + 5

10x + y = 3(x + y)                  Remove the brackets          

10x + y = 3x + 3y                 Substitute the x + 5 into the second equation for y

10x + x + 5 = 3x + 3(x + 5)   Remove the brackets on the right.

10x + x + 5 = 3x + 3x +15     Collect like terms on each side.

11x + 5 = 6x + 15                   Subtract 5 from both sides

11x + 5 - 5 = 6x + 15 - 5        Collect like terms

11x = 6x + 10                         Subtract 6x from both sides

11x - 6x = 6x - 6x + 10

5x =  10                                 Divide by 5

5x/5 = 10/5

x = 2

y = x + 5

y = 2 + 5

y = 7

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3 years ago
What value for x makes the sentence true?<br> 4X-4 = 20
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Step-by-step explanation:

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Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

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.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

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

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

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

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

5 0
3 years ago
Copy the table and fill in the missing values.
Maslowich

Answer:

t53

Step-by-step explanation:

5 0
3 years ago
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