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3 years ago

PLEASE HELP SOLVE!! WILL MARK THE BRAINIES

Mathematics

1 answer:

Vitek1552 [10]3 years ago

7 0

Answer:

Your answer is C.

Step-by-step explanation:

Mark as brainlest this time lol.

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What is the product of 3(-16)?

weeeeeb [17]

Answer: -48

Step-by-step explanation:

16 * 3 = 48

But because the 16 is negative, it becomes -48

4 0

3 years ago

Mr.tom is making 8 of his famous cheese sandwiches.he uses 1/4 of a pound of chesse for each sandwich.

Pavlova-9 [17]

A. Mr. Tom needs 2 pounds of cheese.
1/4 multiplied by 8
b. Mr. Tom will need to buy 3/4 of a pound of cheese.
1/4 multiplied by 3
c. Mr. Tom is correct because to get 1/2 of a sandwich you need 1/2 as much of cheese as needed in 1 whole sandwich. You would multiply 1/2 by 1/4 which equals 1/8. 1/2 of a sandwich requires 1/8 of a pound of cheese.

3 0

3 years ago

A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th

Valentin [98]

Answer:

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

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

The mean calculated for this case is $\bar X=12.8$

The sample deviation calculated $s=4.324 \approx 4.3$

$12.8-4.604\frac{4.324}{\sqrt{5}}=3.896$

$12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704$

So on this case the 99% confidence interval would be given by (3.896;21.704)

Step-by-step explanation:

Data: 10,9,20,13,12

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 $\bar X=12.8$

The sample deviation calculated $s=4.324$

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 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,4)".And we see that $t_{\alpha/2}=4.604$

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

$12.8-4.604\frac{4.324}{\sqrt{5}}=3.896$

$12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704$

So on this case the 99% confidence interval would be given by (3.896;21.704)

6 0

3 years ago

Plzzzzzzzzzzzzzz answer

Aleksandr-060686 [28]

If the amount of weeks is w, then -1.5w is the amount of hair cut off due to that she subtracts 1.5 from her hair every week. Therefore, since the base amount is 11.5, we have to add that to the expression to get 11-1.5w

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4 years ago

Simplifying trigonometric expressions help?

Studentka2010 [4]

7.

$\dfrac{\sin\theta\cdot\cot\theta}{\tan\theta\cdot\cos\theta}=\dfrac{\sin\theta}{\cos\theta}\cdot\dfrac{1}{\tan\theta}\cdot\cot\theta=\tan\theta\cdot\dfrac{1}{\tan\theta}\cdot\cot\theta=\boxed{\cot\theta}$

8.

$\dfrac{\cot\theta}{\cos\theta}\cdot\dfrac{\sin\theta\cdot\cos^2\theta}{1-\cos^2\theta}=\dfrac{\cot\theta}{\cos\theta}\cdot\dfrac{\sin\theta\cdot\cos^2\theta}{\sin^2\theta}=\cot\theta\cdot\dfrac{\cos\theta}{\sin\theta}=\cot\theta\cdot\cot\theta=\\\\\\=\boxed{\cot^2\theta}$

4 0

3 years ago