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2 years ago

Help please, I need it fast

Mathematics

2 answers:

Keith_Richards [23]2 years ago

7 0

Answer:

Step-by-step explanation:

using the rule of exponents

$\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , then

$\frac{2^{-3} }{2^{-5} }$ = $2^{-3-(-5)}$ = $2^{(-3+5)}$ = 2² → A

romanna [79]2 years ago

5 0

Answer:

(d) 1 / 2^8

this is the answer

both equations answers are kind of same, so (d) is the answer for your question

if this helped you please brainliest?

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Jean works for the government and was conducting a survey to determine the income levels of a number of different neighborhoods

Brrunno [24]

Answer:

0.0418 = 4.18% probability that the average income level in the neighborhoods was less than $38,000.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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.

Jean knows that the mean income level in the country is $40,000, with a standard deviation of $2,000.

This means that $\mu = 40000, \sigma = 2000$

Jean selected three neighborhoods and determined the average income level.

This means that $n = 3, s = \frac{2000}{\sqrt{3}} = 1154.7$

What is the probability that the average income level in the neighborhoods was less than $38,000

This is the pvalue of Z when X = 38000. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{38000 - 40000}{1154.7}$

$Z = -1.73$

$Z = -1.73$ has a pvalue of 0.0418

0.0418 = 4.18% probability that the average income level in the neighborhoods was less than $38,000.

7 0

3 years ago

EXPLAIN HOW YOU WOULD USE REPEATED SUBTRACTION TO FIND 8 DIVIDED BY 2

Zepler [3.9K]

Imagine that you have 8 apples which you want to hand out to 2 people. You want each person to get an equal amount of apples, so you decide to give 2 apples at a time (one apple to each), until you run out or you don't have enough to give each of the 2 people another apple. Notice that each time you give out 2 apples, you have to subtract 2 from the total left over, and each person gets one apple. Let's see what happens:
1. 8-2=6 -- each person now has 1 apple, and there are 6 more to hand out;
2. 8-2-2=4 -- each person now has 2 apples, and there are 4 more to pass out;
3. 8-2-2-2=2 -- each person now has 3 apples, and there are 2 more to distribute;
4. 8-2-2-2-2=0 -- each person now has 4 apples and there are no more apples to pass out.
Then you hand out 8 apples to 2 people and each of the get 4 apples. This meansthat 8÷2=4.

6 0

3 years ago

Read 2 more answers

Solve each equation by completing the square. If necessary, round to the nearest hundredth.

Ksenya-84 [330]

$1.)\\ \\ b^2+10b=75\\ \\ b^2+10b-75=0\\ \\\Delta = b^{2}-4ac = 10^{2}-4*1*(75)=100+300=400 \\ \\\sqrt{\Delta }=\sqrt{400}=20$

$b_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-10-20}{2}=\frac{-30}{2}=-15\\ \\b_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{-10+20}{2}=\frac{10}{2}=5$

$2.)\\ \\ n^2-20n=-75\\ \\ n^2-20n+75=0\\ \\\Delta = b^{2}-4ac = (-20)^{2}-4*1*75=400-300=100 \\ \\\sqrt{\Delta }=\sqrt{100}=10$

$n_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{20-10}{2}=\frac{10}{2}=5\\ \\n_{2}=\frac{-b+\sqrt{\Delta }}{2a} \frac{20+10}{2}=\frac{30}{2}=15$

$3.)\\ \\t^2+8t-9=0\\ \\\Delta = b^{2}-4ac = 8^{2}-4*1* (-9)=64+36=100 \\ \\\sqrt{\Delta }=\sqrt{100}=10$

$t_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-8-10}{2}=\frac{-18}{2}=-9\\ \\t_{2}=\frac{-b+\sqrt{\Delta }}{2a} \frac{-8+10}{2}=\frac{2}{2}=1$

$4.)\\ \\m^2-2m-8=0\\ \\\Delta = b^{2}-4ac = (-2)^{2}-4*1* (-8)=4+32=36\\ \\\sqrt{\Delta }=\sqrt{36}=6$

$m_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{2-6}{2}=\frac{-4}{2}=-2\\ \\m_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{2+6}{2}=\frac{8}{2}=4$

$5.)\\ \\ v^2+4v-2=0\\ \\\Delta = b^{2}-4ac = 4^{2}-4*1* (-2)=16+8=24\\ \\\sqrt{\Delta }=\sqrt{24}= \sqrt{4*6}=2\sqrt{6}$

$v_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-4-2\sqrt{6}}{2}=\frac{2(-2-\sqrt{6}}{2}= -2-\sqrt{6}\approx -2-2,45\approx -4,45\\ \\v_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{-4+2\sqrt{6}}{2}=\frac{2(\sqrt{6}-2}{2}= \sqrt{6}-2 \approx 2,45-2\approx 0,45$

3 0

3 years ago

What is halfway between 4 /5 and − 2 /3 ?

babymother [125]

Ans= 11-15

Explanation on next comment

7 0

2 years ago

Can a system of linear equations have exactly three ordered pairs for the solution?

In-s [12.5K]

No, a system of linear equations cannot have exactly three ordered pairs for the solution.

7 0

3 years ago