Two numbers that multiply to 20 and add up to -12

Scores on the SAT Mathematics test are believed to be normally distributed. The scores of a simple random sample of five student

AysviL [449]

Answer:

The mean calculated for this case is $\bar X=584$

And the 95% confidence interval is given by:

$584-2.776\frac{86.776}{\sqrt{5}}=476.271$

$584+2.776\frac{86.776}{\sqrt{5}}=691.729$

So on this case the 95% confidence interval would be given by (476.271;691.729)

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=584$

The sample deviation calculated $s=86.776$

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

$df=n-1=5-1=4$

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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that $t_{\alpha/2}=2.776$

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

$584-2.776\frac{86.776}{\sqrt{5}}=476.271$

$584+2.776\frac{86.776}{\sqrt{5}}=691.729$

So on this case the 95% confidence interval would be given by (476.271;691.729)

3 0

3 years ago

Find the magnitude and direction (in degrees) of the vector. (Assume 0° ≤ < 360°.)

ddd [48]

$6i~~ + ~~2\sqrt{3}j\implies < \stackrel{a}{6}~~,~~\stackrel{b}{2\sqrt{3}} > \\\\[-0.35em] ~\dotfill\\\\ \stackrel{magnitude}{\sqrt{a^2+b^2}}\implies \sqrt{6^2+(2\sqrt{3})^2}\implies \sqrt{36+(2^2\cdot 3)}\implies \sqrt{36+12}\implies \sqrt{48} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{direction}{tan^{-1}\left( \cfrac{b}{a}\right)}\implies tan^{-1}\left( \cfrac{2\sqrt{3}}{6} \right)\implies tan^{-1}\left( \cfrac{\sqrt{3}}{3} \right) \\\\\\ tan^{-1}\left( \cfrac{1}{\sqrt{3}} \right)\implies 30^o$

3 0

2 years ago

A diver went 25.65 feet below the surface of the ocean and then 16.5 feet further down he then rose 12.45 feet write and solve a

Anastaziya [24]

So a diver went 25.65 ft BELOW so it's negative.
then he went 16.5 ft further down, so that's negative also.
he then rose 12.45 ft, so that's positive.
so the expression is: -25.65+-16.5+12.45, the answer is 23.7

8 0

3 years ago

Dan is comparing the cost of having printed and prepared by two companies

Kay [80]

For 4 copies, the total cost for preparing and printing will be the same.

Step-by-step explanation:

Let,

p be the number of pages

Company A;

Charges of preparing = $6

Charges of per page print = $2.50

A(p)= 2.50p + 6 Eqn 1

Company B;

Charges of preparing = $4

Charges of per page print = $3

B(p)= 3p + 4 Eqn 2

For equaling the cost;

A(p) = B(p)

$2.50p+6=3p+4\\2.50p-3p=4-6\\-0.50p=-2$

Dividing both sides by -0.50

$\frac{-0.50p}{-0.50}=\frac{-2}{-0.50}\\p=4$

For 4 copies, the total cost for preparing and printing will be the same.

Keywords: Addition, division

Learn more about division at:

#LearnwithBrainly

8 0

3 years ago

Suppose I am playing a sport that does not allow me to end a match in a tie (AKA I have to either win or lose). If my probabil

Leviafan [203]

Answer: The required probability is 0.3456.

Step-by-step explanation:

Since we have given that

Probability of winning at any given time = 0.6

Probability of losing at any given time = 1-0.6 = 0.4

Number of total matches = 5

Number of won matches = 3

So, using "Binomial distribution", we get that

$P(X=3)=^5C_3(0.6)^3(0.4)^2\\\\P(X=3)=0.3456$

Hence, the required probability is 0.3456.

6 0

3 years ago