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2 years ago

Select all ratios equivalent to 20:5.

Mathematics

2 answers:

Anettt [7]2 years ago

4 0

The answer is : 40:10

*multiply by 2

If helpful, brainliest please !

OLga [1]2 years ago

4 0

Answer:

40:10 & 8:2

Step-by-step explanation:

20 * 2 = 40

5 * 2 = 10

8 * 2.5 = 20

2 * 2.5 = 5

2:1 does not work because only 2 * 10 = 20, but 1 *10 = 10 and there is no 10. There is only 5.

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Find the missing number in the proportion.

Soloha48 [4]

Answer: 7

Step-by-step explanation: Cross multiply, so 15 times x and 21 times 5. 21 times 5 is 105 then you divide by 15 which gives you 7.

8 0

2 years ago

How large a sample should be selected to provide a 95% confidence interval with a margin of error of 6? Assume that the populati

marshall27 [118]

Using the z-distribution, it is found that a sample of 171 should be selected.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

The margin of error is:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

z is the critical value.

n is the sample size.

$\sigma$ is the standard deviation for the population.

For this problem, the parameters are:

$z = 1.96, \sigma = 40, M = 6$

Hence we solve for n to find the needed sample size.

$M = z\frac{\sigma}{\sqrt{n}}$

$6 = 1.96\frac{40}{\sqrt{n}}$

$6\sqrt{n} = 40 \times 1.96$

$\sqrt{n} = \frac{40 \times 1.96}{6}$

$(\sqrt{n})^2 = \left(\frac{40 \times 1.96}{6}\right)^2$

n = 170.7.

Rounding up, a sample of 171 should be selected.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

4 0

2 years ago

PLEASE HELP ME PLEASE

il63 [147K]

Looks like you meant to write (p/q)³ if so it's power of a quotient

3 0

3 years ago

Read 2 more answers

Find the value of x. A.18 B.15 C.16 D.19 PLEASE HELP!!!!!

In-s [12.5K]

Answer:

x = 19

Step-by-step explanation:

123°+3x = 180°

3x = 180°-123°

3x = 57

x = 57/3

x = 19

The reason why we use 123° + 3x = 180° because both are adjacent angles. Adjacent angles add up to 180°

7 0

2 years ago

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago