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trasher [3.6K]
3 years ago
11

Miguel tells his teacher1/5 is the same as 20%. Which best justifies Miguel’s answer? a.5 goes into 100 twenty times, so 20% is

the answer. b. 5 goes into 20 four times, and 5 times 4 is 20, so 20% is the answer. c. 5 goes into 100 twenty times, and 1 times 20 is 20, so 20% is the answer. d. 5 goes into 20 four times, and 5 times 4 times 1 is 20, so 20% is the answer.
Mathematics
2 answers:
svetoff [14.1K]3 years ago
3 0

Answer:

C. 5 goes into 100 twenty times, and 1 times 20 is 20

Step-by-step explanation:

Since, we know that when we multiply both numerator and denominator of a fraction by a same number then we obtain an equivalent fraction,

Here, the given fraction,

\frac{1}{5}

By the above statement,

\frac{1}{5}=\frac{1\times 20}{5\times 20}=\frac{20}{100}

Now,

a\%=\frac{a}{100}

\implies \frac{20}{100}=20\%

Hence,

\frac{1}{5}=20\%

Option C is correct.

Ne4ueva [31]3 years ago
3 0

Answer:

I would say A, but if you look closer and explain C it makes more sense, so it’s C

Step-by-step explanation:

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25 km<br> 7 km<br> 18 km<br> 24 km<br> Find the area!!
Anna35 [415]

Answer:

75600k^4 m^4 Hope that help !

Step-by-step explanation:

1. Take out the constants (25×7×18×24) kkkkmmmm

2. Simplify 25×7=175 (175×18×24)kkkkmmmm

3. Simplify 175×18=3150 (3150×24)kkkkmmmm

4. 3150×24= 75600kkkkmmmm

5. The answer would be 75600k^4 m^4

7 0
3 years ago
3.) What must be the value of b to make the statement true?
cricket20 [7]
Answer= 7
Explanation: combine all like numbers
5-b+4-2=0
5-2+4=7
7-b=0
Then you can tell the answer is 7 because that is the only thing that can replace b and equal 0
7 0
3 years ago
Irrational numbers are imaginary. true or false
Vedmedyk [2.9K]

Imaginary numbers are not rational which may lead you to believe that they must be irrational. Though logical, you would still be incorrect because “irrational” also applies only to real numbers.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
The amount of time all students in a very large undergraduate statistics course take to complete an examination is distributed c
Anestetic [448]

Answer:

a) The mean is \mu = 60

b) The standard deviation is \sigma = 9

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 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.

The probability a student selected at random takes at least 55.50 minutes to complete the examination equals 0.6915.

This means that when X = 55.5, Z has a pvalue of 1 - 0.6915 = 0.3085. This means that when X = 55.5, Z = -0.5

So

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

-0.5 = \frac{55.5 - \mu}{\sigma}

-0.5\sigma = 55.5 - \mu

\mu = 55.5 + 0.5\sigma

The probability a student selected at random takes no more than 71.52 minutes to complete the examination equals 0.8997.

This means that when X = 71.52, Z has a pvalue of 0.8997. This means that when X = 71.52, Z = 1.28

So

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

1.28 = \frac{71.52 - \mu}{\sigma}

1.28\sigma = 71.52 - \mu

\mu = 71.52 - 1.28\sigma

Since we also have that \mu = 55.5 + 0.5\sigma

55.5 + 0.5\sigma = 71.52 - 1.28\sigma

1.78\sigma = 71.52 - 55.5

\sigma = \frac{(71.52 - 55.5)}{1.78}

\sigma = 9

\mu = 55.5 + 0.5\sigma = 55.5 + 0.5*9 = 55.5 + 4.5 = 60

Question

The mean is \mu = 60

The standard deviation is \sigma = 9

6 0
3 years ago
What is the value of C in the matrix equation below?
lara [203]

Given:

The matrix equation is:

\begin{bmatrix}-18&3&5\end{bmatrix}-C=\begin{bmatrix}-22&1&12\end{bmatrix}

To find:

The value of matrix C.

Solution:

Let A=\begin{bmatrix}-18&3&5\end{bmatrix},B=\begin{bmatrix}-22&1&12\end{bmatrix}. Then the given equation can be rewritten as

A-C=B

-C=B-A

C=-(B-A)

C=A-B

On substituting the values of the matrices, we get

C=\begin{bmatrix}-18&3&5\end{bmatrix}-\begin{bmatrix}-22&1&12\end{bmatrix}

C=\begin{bmatrix}-18-(-22)&3-1&5-12\end{bmatrix}

C=\begin{bmatrix}-18+22&2&-7\end{bmatrix}

C=\begin{bmatrix}-4&2&-7\end{bmatrix}

Therefore, the correct option is C.

6 0
3 years ago
Read 2 more answers
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