All categories

4 years ago

LOTS OF POINTS - SIMPLE MATH

Mathematics

1 answer:

aliya0001 [1]4 years ago

5 0

Answer:

$10. \text{Opposite of a Difference} \implies \text{E. } -(8 -3m) = 3m - 8$

$11. \text{Opposite of a Sum} \implies \text{A. } -[(-r) + 2p] = -(-r) -2p$

$12. \text{Opposite of an Opposite} \implies \text{F. }-[-(9 - 2w)] = 9 - 2w$

$13. \text{Multiplication by 0} \implies \text{B. }16d -(3d + 2)(0) = 16d - 0 = 16d$

$\boxed{\text{The product of two numbers is equal to 0} \implies 0 \cdot a = 0}$

$14. \text{Multiplication by -1} \implies \text{D. } -(4r + 3s) + t = (-1)(4r + 3s) + t$

$15. \text{Distributive Property} \implies \text{C. } 5(2-x) = 10-5x$

You might be interested in

The county fair charges $1.25 per ticket for the rides. Jermaine bought 25 tickets for the rides and spent a total of $43.75 at

romanna [79]

Answer:

Ocean currents act as conveyor belts of warm and cold water,

Step-by-step explanation:

sending heat toward the polar regions and helping tropical areas cool off, thus influencing both weather and climate. ... The tropics are particularly rainy because heat absorption, and thus ocean evaporation, is highest in this area.

4 0

4 years ago

Pablo wanted to see if a 12-inch straw would fit inside a small retangular box. He noticed that it only fits diagonally. The box

Artist 52 [7]

Answer:

11.61895003862225

Step-by-step explanation:

7 0

2 years ago

The cost of Mr.pattern car insurance increases by 5% to 86.82 per month.what was the cost of his insurance before it increased?

ohaa [14]

Answer:

82.479

Step-by-step explanation:

You need to find out what 95% of the end cost is to determine the cost before the increase.

you can also find out what 5% per month and then subtract it form 86.82

I hope that makes sense and helps :)

4 0

4 years ago

PLEASEEEE HELPPPP 10th GRADE GEOMETRY!!!

castortr0y [4]

1 = 30 degrees
2 = 30 degrees
l3 = 120 degrees
4 = 60 degrees
5= 60 degrees
6 = 60 degrees
7 = 30 degrees
8 = 120 degrees
9 = 30 degrees

6 0

3 years ago

A fitness center is interested in finding a 90% confidence interval for the mean number of days per week that Americans who are

Blababa [14]

Answer:

A. Normal

B. Between 2.078 and 2.722 visits.

C. About 90 percent of these confidence intervals will contain the true population mean number of visits per week and about 10 percent will not contain the true population mean number of visits per week.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

Question A:

By the Central Limit Theorem, normal.

Question B:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.05 = 0.95$ , so Z = 1.645.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645\frac{2.9}{\sqrt{219}} = 0.322$

The lower end of the interval is the sample mean subtracted by M. So it is 2.4 - 0.322 = 2.078 visits

The upper end of the interval is the sample mean added to M. So it is 2.4 + 0.322 = 2.722 visits.

Between 2.078 and 2.722 visits.

Question c:

90% confidence level, so 90% will contain the true population mean, 100 - 90 = 10% wont.

About 90 percent of these confidence intervals will contain the true population mean number of visits per week and about 10 percent will not contain the true population mean number of visits per week.

5 0

3 years ago