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saul85 [17]
3 years ago
15

Rick ran for hour on Wednesday for a distance of 4.2 miles. On Thursday, he ran for 1 hour at the same speed he ran on Wednesday

. How
many miles did Rick run on Thursday?
A 4.9 5 miles
B.3.15 miles
C5.60 miles
D5.25


Help please
Mathematics
1 answer:
just olya [345]3 years ago
5 0

Answer: C.5.60 miles

Step-by-step explanation:

I don’t know the explanation but I got it correct on a test that I was taking...

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14

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I am confused in these problems ?<br>8. 9.
Liula [17]

Answer:

8. 14

9. 720

Step-by-step explanation:

Both questions are asking you to find the GCD (Greatest Common Divisor) and LCM (Lowest Common Multiple).

8. Find the greatest number that can divide 392 and 462 exactly.

Find the GCD of 392 and 462:

  1. Find the prime factorization of 392

2^3 × 7^2 (2 × 2 × 2 × 7 × 7)

  2. Find the prime factorization of 462

2 × 3 × 7 × 11

  3. To find the GCD, multiply all prime factors common to all numbers  

GCF = 2 × 7, therefore GCD = 14.

9. What is the smallest number that is divisible by 20, 48, and 72?

Find the LCM of 20, 48, and 72:

  1. Find the prime factorization of 20

2^5 (2 × 2 × 5)

  2. Find the prime factorization of 48

2^4 x 3 (2 × 2 × 2 × 2 × 3)

  3. Find the prime factorization of 72

2^3 x 3^2 (2 × 2 × 2 × 3 × 3)

  4. To find the LCM, multiply all prime factors

LCM = 2^4 × 3^2 × 5 (2 × 2 × 2 × 2 × 3 × 3 × 5), therefore GCF = 720

Calculator Input:

The easiest way to find the GCD or LCM of 2 or more numbers is by using a scientific calculator. Input GCD(x,y) or LCM(x,y) for the answer. This will work for any amount of numbers as long as the input is logical.

For questions 8 and 9, you would type in GCD(392,462) and LCM(20,48,72).

Not all calculators are made the same, so other labels might be GCF or HCF instead of GCD, and LCF instead of LCM. I use GCD and LCM because that's how it is on my calculator. There isn't any difference, so using any label is fine.

5 0
2 years ago
Here are summary statistics for randomly selected weights of newborn​ girls: n=185​,
Anna11 [10]

Answer:

The confidence  interval for the population mean μ​ is 32.487

Step-by-step explanation:

Given :

Number of weights of newborn​ girls n=185.

Mean \bar{x}=31.4 hg

Standard deviation s=7.5 hg

Use a 95​% confidence level i.e. cl=0.95

To find : What is the confidence interval for the population mean μ​?

Solution :

Using t-distribution,

The degree of freedom DF=n-1=185-1=184

\alpha=\frac{1-0.95}{2}

\alpha=0.025

The t critical value is t=1.973.

The confidence interval build is

CI=\bar{x}\pm(t\cdot \frac{s}{\sqrt{n}})

Substitute the values,

CI=31.4\pm(1.973\cdot \frac{7.5}{\sqrt{185}})

CI=31.4\pm(1.973\cdot 0.5514)

CI=31.4\pm(1.087)

31.4+1.087

32.487

Therefore, the confidence  interval for the population mean μ​ is 32.487

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3 years ago
Which is a conditional statement for this sentence. All birds can fly
Makovka662 [10]
All birds, but penguins can fly
7 0
3 years ago
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