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2 years ago

Compare 3.5 • 10^4 to standard form

Mathematics

1 answer:

GREYUIT [131]2 years ago

7 0

Answer:

35,000

Step-by-step explanation:

^4 means 4 zeros

10^4 = 10,000

3.5 times 10,000 =

35,000

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Determine the value of X that makes the following equation true

valentinak56 [21]

What you want to do is cross multiply so you get 5(x+4)=7(3x+28) then you distrubute the 5 and then you distribute the 5 and the 7 to the number in the parentheses 5x+20=21x+196
subtract 5x on both sides to get 20=16x+196 then you subtract 196 on both sides to get -176=16x last what you want to do is divide 16 on both sides to get x=-11

5 0

3 years ago

Student records suggest that the population of students spends an average of 6.30 hours per week playing organized sports. The p

Ymorist [56]

Answer:

a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

Step-by-step explanation:

To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem

Normal probability distribution

Problems of normally distributed samples 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

In this problem, we have that:

$\mu = 6.3, \sigma = 2.1, n = 49, s = \frac{2.1}{\sqrt{49}} = 0.3$

A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?

This is the pvalue of Z when X = 7.1 subtracted by the pvalue of Z when X = 5.5.

By the Central Limit Theorem, the formula for Z is:

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

X = 7.1

$Z = \frac{7.1 - 6.3}{0.3}$

$Z = 2.67$

$Z = 2.67$ has a pvalue of 0.9962

X = 5.5

$Z = \frac{5.5 - 6.3}{0.3}$

$Z = -2.67$

$Z = -2.67$ has a pvalue of 0.0038

So there is a 0.9962 - 0.0038 = 0.9924 = 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.

This is the pvalue of Z when X = 6.7 subtracted by the pvalue of Z when X = 5.9

X = 6.7

$Z = \frac{6.7 - 6.3}{0.3}$

$Z = 1.33$

$Z = 1.33$ has a pvalue of 0.9082

X = 5.9

$Z = \frac{5.9 - 6.3}{0.3}$

$Z = -1.33$

$Z = -1.33$ has a pvalue of 0.0918.

So there is a 0.9082 - 0.0918 = 0.8164 = 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

5 0

3 years ago

Whats is the anwser to 469 plus 8346

NeX [460]

469 + 8346 = 8,815...is that what you needed to know?

3 0

3 years ago

Nicole deposited $2,000 at 6% simple interest. How long will it be before she has $2,600 in her account?

coldgirl [10]

$\bf \qquad \textit{Simple Interest Earned Amount}\\\\
A=P(1+rt)\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &2,600\\
P=\textit{original amount deposited}\to& \$2,000\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
t=years
\end{cases}$

solve for "t"

7 0

3 years ago

a shopkeeper sold a pen at loss of 20%. if he had sold it for rs 10 more, he would have gain 5%.find the cost price of pen

AlexFokin [52]

Rs 200
since the questions states that if he sold it for 10 more, he would have gained 5%, 10=5%
1%=2
100%=200

8 0

3 years ago