I need help here with explanation pleaseeeee

You choose a marble at random from a bag containing 8 brown marbles 6 yellow marbles and one purple marble. You replace the marb

Andrej [43]

Pick 1: 6 yellow out of 15 total marbles = 6/15 = 2/5
Pick 2: 6 yellow out of 15 total marbles = 6/15 = 2/5

Pick 1 AND Pick 2
2/5 x 2/5 = 4/25

5 0

3 years ago

3l of ice cream is 10$ 4l of ice cream is 12% which is the better choice

Korolek [52]

Answer:

4l of icecream

Step-by-step explanation:

10÷3=3.33

12÷4=3

3 0

2 years ago

Please help I give lots of points.

Minchanka [31]

Answer:

I believe your answer is z5√z the fourth option

7 0

4 years ago

Car rentals involve a $130 flat fee and an additional cost of $31.67 a day what is the maximum number of days you can rent a car

Dmitry_Shevchenko [17]

The equation is $130 + $31.67x = 500
Subtract 130 from both sides $31.67x = $370
Divide $31.67 from both sides
X=11.68
BUT
you can’t have .68 of a day so the answer is 11 days.

8 0

4 years ago

Read 2 more answers

A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$

And replacing we got:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

5 0

4 years ago