Please help with this!

Simplify (7 – c)(–1)

Shkiper50 [21]

Answer:

-7 + C in other form, C + 7

Just open the bracket

4 0

2 years ago

Read 2 more answers

Help would be greatly appreciated!

eimsori [14]

b

Take a look at everything inside the brackets sqrt(50x^2) = sqrt(5*5*x*x * 2) For ever 2 factors you can bring one out and drop the other one. That means take out 5 * x * sqrt(2). Two is what is left inside the brackets. B must = 2.

c

Do the same thing here. Write all the primes under the square root. Take out 1 for every two under the root sign. sqrt(32x) = sqrt(2*2*2*2*2*x) You can bring out two 2s. There is one left over. Leave it under the root sign. The x is a loner. It stays under the root sign. c = 2 * 2 = 4.

e

Again do the prime factor thing. sqrt(18n) = sqrt(3*3*2*n) = 3*sqrt(2*n)

e = 3

g

sqrt(72*x*x) = sqrt(3 * 3* 2 * 2 * 2 * x * x) = 3 *2 * x* sqrt(2) For every 2 prime factors you can pull out 1 of them outside the square root sign. g = 6

Comment

There are many people on the net and on Brainly that will say that you should know the perfect squares from 1 to 100 (say) so 4 9 16 25 36 49 64 81 100 are the numbers that you should memorize. When they are under the root sign, their roots can be taken out as 2 3 4 5 6 7 8 9 10. For this question I think it is better to use the pairs rule I've given you above. If someone else answers they are likely to do it the way it is written up in this paragraph. It's a free country. You are free to take the answer you like best.

7 0

2 years ago

326 ÷ 53 <br><br> Please help me

nataly862011 [7]

The answer to this question is 6.150943396226415

8 0

3 years ago

Benford's Law states that the probability that the first decimal-digit of a raw data sample (from 1 to 9) is given

patriot [66]

Answer: P(4) = 0.097

$\bold{\sum_{m=1}^9=1}$

<u>Step-by-step explanation:</u>

$P(m)=\log(m+1)-\log(m)\\\\.\qquad =\log\bigg(\dfrac{m+1}{m}\bigg)\qquad \text{(using rules for condensing logs)}\\\\\\P(4)=\log\bigg(\dfrac{4+1}{4}\bigg)\\\\\\.\qquad =\log\bigg(\dfrac{5}{4}\bigg)\\\\\\.\qquad =\large\boxed{0.097}$

$\sum_{m=1}^9\log\bigg(\dfrac{m+1}{m}\bigg)\\\\\\.\qquad =\log\bigg(\dfrac{2}{1}\bigg)+\log\bigg(\dfrac{3}{2}\bigg)+...\log\bigg(\dfrac{9}{8}\bigg)+\log\bigg(\dfrac{10}{9}\bigg)\\\\\\.\qquad =\log\bigg(\dfrac{2}{1}\bigg)\bigg(\dfrac{3}{2}\bigg)...\bigg(\dfrac{9}{8}\bigg)\bigg(\dfrac{10}{9}\bigg)\qquad \text{(using rules for condensing logs)}\\\\\\.\qquad =\log\bigg(\dfrac{10!}{9!}\bigg)\\\\\\.\qquad =\log(10)\\\\\\.\qquad =\large\boxed{1}$

8 0

2 years ago

In a random sample of 81 audited estate tax returns, it was determined that the mean amount of additional tax owed was $3408 w

pav-90 [236]

Answer:

B. One can be 90% confident that the mean additional tax owed is between the lower and upper bounds.

Step-by-step explanation:

Given:

n= 81

$\bar{x}=3408$

$\sigma= 2565$

Solution:

A confidence interval, in statistics, refers to the probability that a population parameter will fall between two set values for a certain proportion of times. Confidence intervals measure the degree of uncertainty or certainty in a sampling method. A confidence interval can take any number of probabilities, with the most common being a 95% or 99% confidence level.

Confidence interval = $\bar{x} \pm z * \frac{\sigma}{\sqrt{n}}$