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3 years ago

R(x)=2√x s(x)=√x (rs)(4) = and (r/s)(3)=

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

6 0

(rs)(4)=(2rootX)(rootX)
(rs)(4)=2x
(rs)(4)=8

(r/s)(3)=(2rootX)/(rootX)
(r/s)(3)=2

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wel

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3 years ago

What is the slope of the line that passes through the points (13,-8) and (-11,4)

pogonyaev

Answer:

- $\frac{1}{2}$

Step-by-step explanation:

Find slope using the slope formula : $\frac{y2-y1}{x2-x1}$

Plug in the given points : $\frac{4- (-8)}{-11-13}$

Add the numbers : $\frac{12}{-11-13}$

Calculate the difference : $\frac{12}{-24}$

Reduce the fraction : - $\frac{1}{2}$

Solution : - $\frac{1}{2}$

3 0

3 years ago

Which is larger 143.0 millimeters or 143.0 decimeters?

Natali5045456 [20]

1 millimeter = 10⁻³ m
1 decimeter = 10⁻¹ m
So,decimeters are larger units, and 143.0 decimeters > 143.0 millimeters .

6 0

3 years ago

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 $t_{\alpha/2}=2.101$