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puteri [66]
3 years ago
6

Does anyone understand how to do this? Solve r4<−5 or −2r−7 ≤ 3 . The solution is _or_

Mathematics
1 answer:
blagie [28]3 years ago
5 0
I believe this is the answer

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Caroline served a ball over the net 16 out of 20 times.What percentage did she not make her serve over the net
kow [346]
16-20= 4 so 4/20 is 0.2 or 20%. To get percentage move decimal to the right two times :)
5 0
3 years ago
Read 2 more answers
I need help! Question is in the picture!
Pavlova-9 [17]

Answer: A

Step-by-step explanation: I chose A because I did 11+2x4 and got 19 and then for 22 I did 22+2x4 which is 30.

4 0
3 years ago
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Find the GCF and LCM of each number using any of the methods.​
vodka [1.7K]

The GCF is the smallest shared prime factor for each term.

The LCM is the smallest whole number that both numbers share as multiples.

For example, take #16.

Factors of 25: 1 , 5, 25

Factors of 75: 1 , 3, 5, 15, 25 ,75

The GCF that is shared would then be 25.

The smallest multiple that both share as multiples:

Multiples of 25: 25, 50, 75, etc.

Multiples of 75: 75, etc.

The LCM that is shared would then be 75.

16) 25 and 75

GCF: 25

LCM: 75

17) 35 and 5

GCF: 5

LCM: 35

18) 27 and 36

GCF: 9

LCM: 108

19) 21 and 56

GCF: 7

LCM: 168

20) 54 and 72

GCF: 18

LCM: 216

8 0
2 years ago
If multiplying by 1/4 makes a positive number smaller, then what dividing by 1/4 do to the value of the number? Explain your rea
mars1129 [50]

Answer:

It would quadruple it.

Step-by-step explanation:

When you divide by a fraction you first flip it then you multiply straight across. For example 3 / (1/4) you flip then multiply which gives you 3/1 * 4/1 = 12/1 which simplifies to 12

6 0
3 years ago
Mathematical Statistics with Applications Help Homework
Mars2501 [29]

7.43: Let X denote the random variable for height and \overline X for the sample mean. Then if \mu is the mean of

So the probability that the difference between the sample and population means does not exceed 0.5 inch is

P(|\overline X-\mu|\le0.5) = P\left(\left|\dfrac{\overline X-\mu}{0.25}\right|\le\dfrac{0.5}{0.25}\right) = P(|Z|\le 2) \approx \boxed{0.95}

per the empirical or 68/95/99.7 rule.

7.44: For a sample of size <em>n</em>, the sample standard deviation would be \frac{2.5}{\sqrt n}. We want to find <em>n</em> such that

P(|\overline X-\mu| < 0.4) = 0.95

Comparing to the equation from the previous part, this means we would need

\dfrac{0.4}{\frac{2.5}{\sqrt{n}}} = 0.16\sqrt n = 2 \implies \sqrt n = 12.5 \implies n = 156.25

so a sample of at least 157 men would be sufficient.

7 0
3 years ago
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