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timama [110]
2 years ago
13

Cions: Write the decimal as a percent and as a fraction or mixed number in simplest

Mathematics
1 answer:
Helga [31]2 years ago
7 0
Answer
15%
90%
2%
Explanation
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15/20 times 1/4 thank me if u like brainly
MaRussiya [10]

Answer: 3/16

Step-by-step explanation: To multiply fractions, first multiply across the numerators and then multiply across the denominators.

\frac{15}{20} × \frac{1}{4} = ?

15 × 1 = 15

20 × 4 = 80

\frac{15}{20} can be reduced by dividing both the numerator and denominator by the Greatest Common Factor which is 5.

15 ÷ 5 = 3

80 ÷ 5 = 16

Therefore, \frac{15}{20} × \frac{1}{4} = \frac{3}{16}

3 0
3 years ago
One hundred million divided by ten thousand equals?
SVETLANKA909090 [29]
It equals 10 000
10 000 ×10 000=100 000 000
4 0
3 years ago
Read 2 more answers
14. Which of the following is not a true statement about the scatter plot above?
harina [27]

Answer:

Step-by-step explanation:

5 0
2 years ago
What is a counterexample?◑〰️◑ please help
Strike441 [17]

Answer:

A counterexample is a specific case which shows that a general statement is false. Example 1: Provide a counterexample to show that the statement. "Every quadrilateral has at least two congruent sides" is not always true.

7 0
3 years ago
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The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78
____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

m-t_{a/2,n-1}\frac{s}{\sqrt{n} } ≤ μ ≤ m+t_{a/2,n-1}\frac{s}{\sqrt{n} }

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and t_{a/2,n-1} is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and t_{0.05,9} by 1.8331, we get that the 90% confidence interval for the mean speed is:

74.2-(1.8331)\frac{5.3083}{\sqrt{10} } ≤ μ ≤ 74.2+(1.8331)\frac{5.3083}{\sqrt{10} }

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0
3 years ago
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