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Grace [21]
3 years ago
15

Explain why two thirds is not unit fraction

Mathematics
1 answer:
vodomira [7]3 years ago
8 0
I'm pretty sure it has to have 1 over any number
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Expanded form of 412.638
Darya [45]
(4 x 100) + (1 x 10) + (2 x 1) + (6 x 0.1) + (3 x 0.01) + (8 x 0.001)
7 0
3 years ago
Read 2 more answers
Your family goes to a Southern-style restaurant for dinner. There are 6 people in your family.
swat32

Answer:

We conclude that:

  • The number of people who ordered the chicken dinner = 1
  • The number of people who ordered the steak dinner = 5

Step-by-step explanation:

Given that

  • Some order the chicken dinner for $14
  • some order the steak dinner for $17.
  • Total bill = $99

  • Let 'n' be the number of people who ordered the chicken dinner.
  • Let '6-n' be the number of people who ordered the steak dinner.

Thus, the equation becomes

14n + 17(6-n) = 99

14n + 102 - 17n = 99

-3n = 99 - 102

-3n = -3

Divide both sides by -3

n = 1

and

6-n = 6-1 = 5

Therefore, we conclude that:

  • The number of people who ordered the chicken dinner = 1
  • The number of people who ordered the steak dinner = 5
7 0
2 years ago
the combined area of two squares is 45 square cm. each side one square is twice as long as a side of the other Square. what is t
anygoal [31]
6 centimeters



x will represent the side length of the smaller square, so 2x is the side length of the bigger square.

x*x + 2x*2x = 45

x^2 + 4x^2 = 45

5x^2 = 45

x^2 = 9 (Divide by 5)

sqrt(x^2) = sqrt(9)

x = 3

2x = 6 (Multiply by 2)



Please consider marking this answer as Brainliest to help me advance.
4 0
3 years ago
Help please it’s for home work
Dmitry [639]
Slope would be 2
Y intercept would be 2
y = 2x + 2
6 0
3 years ago
Weights of American adults are normally distributed with a mean of 180 pounds and a standard deviation of 8 pounds. What is the
ahrayia [7]

Answer:

15.87% probability that a randomly selected individual will be between 185 and 190 pounds

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 = 180, \sigma = 8

What is the probability that a randomly selected individual will be between 185 and 190 pounds?

This probability is the pvalue of Z when X = 190 subtracted by the pvalue of Z when X = 185. So

X = 190

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

Z = \frac{190 - 180}{8}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

X = 185

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

Z = \frac{185 - 180}{8}

Z = 0.63

Z = 0.63 has a pvalue of 0.7357

0.8944 - 0.7357 = 0.1587

15.87% probability that a randomly selected individual will be between 185 and 190 pounds

3 0
3 years ago
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