A rectangular room has a perimeter of 70m and is 20m long write an equation to represent the area

(4*10^8)^ simplify expression write in scientific notation

kotegsom [21]

Question asked:
$(4*10^8)^2$

Solve:
=160000000000000000

Turn to scientificnoation:
1.6 * 10^17
Final answer: A 1.6*10^7

6 0

4 years ago

Suppose μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of bra

Stella [2.4K]

Answer:

The 95% confidence interval would be given by $-23.44 \leq \mu_1 -\mu_2 \leq -8.16$

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".

$\bar X_1 =114.1$ represent the sample mean 1

$\bar X_2 =129.9$ represent the sample mean 2

n1=5 represent the sample 1 size

n2=2 represent the sample 2 size

$s_1 =5.08$ sample standard deviation for sample 1

$s_2 =5.37$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =114.1-129.9=-15.8$

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

$df=n_1 +n_2 -1=5+5-2=8$

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(0.025,8)".And we see that $t_{\alpha/2}=\pm 2.31$

The standard error is given by the following formula:

$SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=3.306$

Confidence interval

Now we have everything in order to replace into formula (1):

$-15.8-2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-23.437$

$-15.8+2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-8.163$

So on this case the 95% confidence interval would be given by $-23.44 \leq \mu_1 -\mu_2 \leq -8.16$

8 0

3 years ago

Probabilidad un cubo numérico

yKpoI14uk [10]

Answer:

87

Step-by-step explanation:

87

6 0

3 years ago

The heights of the starting players on each of two basketball teams are shown in the table below.

zloy xaker [14]

Answer:

whats the question? please explain

5 0

2 years ago

During a catered lunch an average of 4 cups of tea are poured per minute. This lunch will last 2 hours. How many gallons of tea

zzz [600]

4 cups are poured every minute...there are 60 minutes in 1 hr...so for 2 hrs, there is 120 minutes...at 4 per minute = 120 * 4 = 480 cups in 2 hrs

16 cups = 1 gal
480 cups = 480/16 = 30 gallons <===

4 0

3 years ago