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3 years ago

Hello are you there my i please have some help

Mathematics

1 answer:

Katarina [22]3 years ago

3 0

Answer:

3^-5

Step-by-step explanation:

x^-3=1/x^3

this rule applies to all negative exponents lol

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(X+m)(c+n) if m=5 and n=-7

oksian1 [2.3K]

Answer: -7/5

Step-by-step explanation:

x = - 5cm^2 - 7m^2 + 35/ m (5c - 7), C = (cross out the equal sign) 7/5 and m = (cross out the equal sign) 0 and n = -7/5

5 0

2 years ago

Write √ 3 x √ 6 in the form b √2 where b is an integer.

tigry1 [53]

Answer:

Step-by-step explanation:

The catch here is to break up sqrt(6) into 2 parts that use primes to define sqrt(6)

sqrt(3)* sqrt(6)

sqrt(6) can be broken up into sqrt(2*3) which equals sqrt(2)*sqrt(3)

sqrt(3)*sqrt(2)(sqrt(3)

sqrt(3)*sqrt(3)*sqrt(2)

3*sqrt(2)

5 0

3 years ago

Help me please will mark

yuradex [85]

Answer:

16 = 216, 17 = 180, 18 = 252

Step-by-step explanation:

Hope this helps!

4 0

3 years ago

Read 2 more answers

Suppose you have a bag containing 2 black marbles and 3 red marbles. You reach into the bag, select a marble, see what color it

sweet [91]

(3/5)(2/5)=6/25
.............................

4 0

3 years ago

There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

7 0

3 years ago