HELP PLEASE

Divide the fractions. thank ya'll super much. (lol) =P

Darina [25.2K]

Answer:

1 4/21

Step-by-step explanation:

5/6 * 10/7

5/3 * 5/7

25/21

1 4/21

8 0

3 years ago

Read 2 more answers

Odysseus found that his troubles varied directly as his distance from his home island of Ithaca. If he had 20 troubles when he w

Vitek1552 [10]

Lol

trouble varies directly as distance
lets say t=trouble and d=distance
t=kd
k is constant

given
when t=20, and d=400
find k
20=400k
divide by 400 both sides
1/20=k

t=(1/20)d

given, d=60
find t

t=(1/20)60
t=60/20
t=3

3 troubles

5 0

2 years ago

Using the simple random sample of weights of women from a data set, we obtain these sample statistics: nequals45 and x overbar

Tomtit [17]

Answer: (141.1, 156.48)

Step-by-step explanation:

Given sample statistics : $n=45$

$\overline{x}=148.79\text{ lb}$

$\sigma=31.37\text{ lb}$

a) We know that the best point estimate of the population mean is the sample mean.

Therefore, the best point estimate of the mean weight of all women = $\mu=148.79\text{ lb}$

b) The confidence interval for the population mean is given by :-

$\mu\ \pm E$ , where E is the margin of error.

Formula for Margin of error :-

$z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given : Significance level : $\alpha=1-0.90=0.1$

Critical value : $z_{\alpha/2}=z_{0.05}=\pm1.645$

Margin of error : $E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69$

Now, the 90% confidence interval for the population mean will be :-

$148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)$

Hence, the 90% confidence interval estimate of the mean weight of all women= (141.1, 156.48)

3 0

3 years ago

A Simple random sample of 100 8th graders at a large suburban middle school indicated that 84% of them are involved with some ty

Setler [38]

Answer: a) (0.755, 0.925)

Step-by-step explanation:

Let p be the population proportion of 8th graders are involved with some type of after school activity.

As per given , we have

n= 100

sample proportion: $\hat{p}=0.84$

Significance level : $\alpha= 1-0.98=0.02$

Critical z-value : $z_{\alpha/2}=2.33$ (using z-value table)

Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

i.e. $0.84\pm (2.33)\sqrt{\dfrac{0.84(1-0.84)}{100}}$

i.e. $\approx0.84\pm 0.085$

i.e. $(0.84- 0.085,\ 0.84+ 0.085)=(0.755,\ 0.925)$

Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)

3 0

3 years ago

While shopping for a speaker, you see a discount.

harina [27]

Answer:

3*4. and 0.3*42.00 and 0.03*42

Step-by-step explanation:

4 0

2 years ago