Please help please i dont know how to do it I need to get my grades up asap

Combine as indicated by the signs. 4/ y^2-9 + 5/y+3

natka813 [3]

$\frac{4}{y^{2}-9 } + \frac{5}{y+3} = \frac{4}{(y-3)(y+3) } + \frac{5}{y+3}= \frac{4}{(y-3)(y+3) } + \frac{5(y-3)}{(y-3)(y+3)}=
\\ \\=\frac{4}{(y-3)(y+3) } + \frac{5y-15}{(y-3)(y+3)}= \frac{5y-11}{(y-3)(y+3)}=\frac{5y-11}{y^{2}-9}
$

8 0

3 years ago

Solve the equation 4=t/2.5 t=

julia-pushkina [17]

Answer:

10

Step-by-step explanation:

Multiply 4 by 2.5 to get 10

8 0

3 years ago

Read 2 more answers

Suppose that the ages of members in a large billiards league have a known standard deviation of σ = 12 σ=12sigma, equals, 12 yea

asambeis [7]

Answer:

$n\geq 23$

Step-by-step explanation:

-For a known standard deviation, the sample size for a desired margin of error is calculated using the formula:

$n\geq (\frac{z\sigma}{ME})^2$

Where:

$\sigma$ is the standard deviation
$ME$ is the desired margin of error.

We substitute our given values to calculate the sample size:

$n\geq (\frac{z\sigma}{ME})^2\\\\\geq (\frac{1.96\times 12}{5})^2\\\\\geq 22.13\approx23$

Hence, the smallest desired sample size is 23

3 0

3 years ago

Can someone explain how to do #3?

dsp73

$\displaysyle (\sqrt 2)^{3}=(2^{\frac{1}{2}})^3=2^{\frac{1}{2} \cdot 3}=2^{\frac{3}{2}}=\sqrt {2^3}=\sqrt 8$

√8 is between 2 and 3, because 2²=4<8, but 3²=9>8. Also, our value is closer to 3 than to 2, so it is more than 2.5 and we have C and D options left.

Among these two numbers we find the one which is closer to √8.

C. 27=√729 ⇒ 2.7=√7.29

D. 28=√784 ⇒ <u>2.8=√7.84</u>

Hence our answer is D) 2.8

5 0

3 years ago

Read 2 more answers

Suppose a continuous probability distribution has an average of μ=35 and a standard deviation of σ=16. Draw 100 times at random

yulyashka [42]

Answer:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

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 normally distributed samples are 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.

For sums, we have that the mean is $\mu*n$ and the standard deviation is $s = \sigma \sqrt{n}$

In this problem, we have that:

$\mu = 100*35 = 3500, \sigma = \sqrt{100}*16 = 160$

This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.

X = 4000

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

By the Central Limit Theorem

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

$Z = \frac{4000 - 3500}{160}$

$Z = 3.13$

$Z = 3.13$ has a pvalue of 0.9991

X = 3000

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

$Z = \frac{3000 - 3500}{160}$

$Z = -3.13$

$Z = -3.13$ has a pvalue of 0.0009

0.9991 - 0.0009 = 0.9982

So the correct answer is:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

5 0

3 years ago