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Vikki [24]
3 years ago
6

PLEASE HELP ME I DO'T UNDERSTAND DIS MATh

Mathematics
1 answer:
sladkih [1.3K]3 years ago
3 0

Answer:

60%

Step-by-step explanation:

A percent is technically just a fraction over 100.

To get that here, we simply divide the part (which is 1560 people) by the whole (which is 2600 people) and then multiply that by 100:

(1560 / 2600) * 100 = 0.6 * 100 = 60/100

Thus, we know that the percent is 60%.

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There are three fourths in six eights.
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Solve T=L(7 + RS) for S.<br> Please help
nikklg [1K]

multiply the ones in bracket by the L

T = 7× L + L × RS

T = 7L + LRS

making S the subject by transferring 7L to the left hand side of the equation.

thus

T - 7L = LRS

dividing through by LR

\frac{T}{LR}  -  \frac{7L}{LR}  = S

S=  \frac{T}{LR}  -  \frac{7}{R}

7 0
3 years ago
Weight gain during pregnancy. In 2004, the state of North Carolina released to the public a large data set containing informatio
pychu [463]

Answer:

1. B. H0: μ1−μ2=0, HA: μ1−μ2≠0

2. z=1.2114

3. P-value=0.2257

4. Do not reject H0

Step-by-step explanation:

We have to perfomr an hypothesis test to see if there is strong evidence that there is a significant difference between the two population means.

The null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2\neq0

Being μ1 the mean average gain for younger mothers and μ2 the mean average gain for mature mothers.

(NOTE: we are comparing means, not proportions, as it is the random variable is the weight gain).

As we are claiming "strong evidence", the level of significance will be 0.01.

For younger mothers, the sample size is n1=840, the sample mean is 30.7 and  the sample standard deviation is s1=14.91.

For mature mothers, the sample size is n2=132, the sample mean is 29.15 and the sample standard deviation is s2=13.46.

The difference between means is

M_d=\mu_1-\mu_2=30.7-29.15=1.55

The standard error of the difference between means is

s_M=\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}=\sqrt{\dfrac{14.91^2}{840}+\dfrac{13.46^2}{132}}=\sqrt{ 0.2647+1.3725}=\sqrt{1.6372}\\\\\\s_M=1.2795

Then, the statistic can be calculated as:

z=\dfrac{M_d-(\mu_1-\mu_2)}{s_M}=\dfrac{1.55-0}{1.2795}=1.2114

The P-value for this z-statistic in a two tailed test is:

P-value=2P(z>1.2114)=0.2257

As the P-value is greater than the significance level, the null hypothesis failed to be rejected.

There is no enough evidence to claim that the real average weight gain differs from mature and youger mothers.

5 0
4 years ago
How to cross multiple 3/5 by 10
postnew [5]

3/5 by 10  = 6

\frac{3}{5} *10 = 6

6 0
3 years ago
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Answer:

what is the question?

Step-by-step explanation:

i don't see a problem or answer choices

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