Find the zeros of the function; To the nearest Hundredth. PLEASE HELP

(PLEASE HELP)<br> Find measurement for angles for 1 and 2

katovenus [111]

Answer:

Angle 1 is = 54° ( vertically opposite angles)

Angle 2 is = Angle 1 (corresponding angles are equal)

therefore Angle 2 = 54°

7 0

3 years ago

If 3x - 6 = 21, what is the value of x - 2 ?

lora16 [44]

3x-6=21
Or,3x=27
Or,x=27/3
Or,x=9
So,Now,
X-2=9-2
=7
Hope it helps

5 0

3 years ago

the set below only contains which types of numbers {-1,5,1/2,15,3.75,36,square root 81,100 are they irrational numbers rational

nika2105 [10]

Answer:

I believe Irrational numbers, That's the answer I used and got it correct!

Step-by-step explanation:

5 0

3 years ago

Andrew plans to retire in 32 years. He plans to invest part of his retirement funds in stocks, so he seeks out information on pa

Debora [2.8K]

Answer:

a) 0.0885 = 8.85% probability that the mean annual return on common stocks over the next 40 years will exceed 13%.

b) 0.4129 = 41.29% probability that the mean return will be less than 8%

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 z-score 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 p-value, 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.

Mean 8.7% and standard deviation 20.2%.

This means that $\mu = 8.7, \sigma = 20.2$

40 years:

This means that $n = 40, s = \frac{20.2}{\sqrt{40}}$

(a) What is the probability (assuming that the past pattern of variation continues) that the mean annual return on common stocks over the next 40 years will exceed 13%?

This is 1 subtracted by the pvalue of Z when X = 13. So

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

By the Central Limit Theorem

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

$Z = \frac{13 - 8.7}{\frac{20.2}{\sqrt{40}}}$

$Z = 1.35$

$Z = 1.35$ has a pvalue of 0.9115

1 - 0.9115 = 0.0885

0.0885 = 8.85% probability that the mean annual return on common stocks over the next 40 years will exceed 13%.

(b) What is the probability that the mean return will be less than 8%?

This is the pvalue of Z when X = 8. So

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

$Z = \frac{8 - 8.7}{\frac{20.2}{\sqrt{40}}}$

$Z = -0.22$

$Z = -0.22$ has a pvalue of 0.4129

0.4129 = 41.29% probability that the mean return will be less than 8%

8 0

2 years ago

$13,957 is invested, part at 7% and the rest at 6%. If the interest earned from the amount invested at 7% exceeds the interest e

ch4aika [34]

Answer:

The Amount invested at 7% interest is $12,855

The Amount invested at 6% interest = $1,102

Step-by-step explanation:

Given as :

The Total money invested = $13,957

Let The money invested at 7% = $p_1$ = $A

And The money invested at 6% = $p_2$ = $13957 - $A

Let The interest earn at 7% = $I_1$

And The interest earn at 6% = $I_2$

$I_1$ - $I_2$ = $833.73

Let The time period = 1 year

Now,<u> From Simple Interest method</u>

Simple Interest = $\dfrac{\textrm principal\times \textrm rate\times \textrm time}{100}$

Or, $I_1$ = $\dfrac{\textrm p_1\times \textrm 7\times \textrm 1}{100}$

Or, $I_1$ = $\dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}$

And

$I_2$ = $\dfrac{\textrm p_2\times \textrm 6\times \textrm 1}{100}$

Or, $I_2$ = $\dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}$

∵ $I_1$ - $I_2$ = $833.73

So, $\dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}$ - $\dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}$ = $833.73

Or, 7 A - 6 (13,957 - A) = $833.73 × 100

Or, 7 A - $83,742 + 6 A = $83373

Or, 13 A = $83373 + $83742

Or, 13 A = $167,115

∴ A = $\dfrac{167115}{13}$

i.e A = $12,855

So, The Amount invested at 7% interest = A = $12,855

And The Amount invested at 6% interest = ($13,957 - A) = $13,957 - $12,855

I.e The Amount invested at 6% interest = $1,102

Hence,The Amount invested at 7% interest is $12,855

And The Amount invested at 6% interest = $1,102 . Answer

8 0

3 years ago