Ben is 444 times as old as Ishaan. 666 years ago, Ben was 666 times as old as Ishaan.

A worker was paid a salary of $10,500 in 1985. Each year, a salary increase of 6% of the previous year's salary was awarded. How

Mazyrski [523]

Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.

Here is how much the worker earned each year:

In the year 1985 the worker earned $10,500.

In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:

$10,500(1+0.06).

In the year 1987 the worker earned

$10,500(1+0.06) + 0.06[$10,500(1+0.06)].

Now we can factorize $10,500(1+0.06) and write the earnings as:

$10,500(1+0.06) [1+0.06]= $$10,500(1.06)^2$ .

Similarly we can check that in the year 1987 the worker earned $$10,500(1.06)^3$ , which makes the pattern clear.

We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:

$10,500+10,500(1.06)^1+10,500(1.06)^2+10,500(1.06)^3...10,500(1.06)^{26}$ .

Factorizing, we have

$=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]$

We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula

$\sum_{i=1}^{n} a_i= a(\frac{1-r^n}{1-r})$ (where a is the first term and r is the common ratio) we have:

$\sum_{i=1}^{26} a_i= 1(\frac{1-(1.06)^{26}}{1-1.06})= \frac{1-4.55}{-0.06}= 59.17$ .

Finally, multiplying 10,500 by 59.17 we have 621.285 ($).

The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating $(1.06)^{26}$ .

Answer: D

4 0

3 years ago

Alejandro is flying a kite, holding his hands a distance of 3.5 feet above the ground

Lemur [1.5K]

Answer: 52.34

Step-by-step explanation: sin23=opposite/hypotenuse=x/125

sin23/1=x/125

125sin23=x

x=48.841391... add 3.5 = 52.341391... round to 52.34

7 0

3 years ago

HELP PLEASEWhat is the value of the exponential expression below?

vladimir2022 [97]

Hi, I got C because you just put 1/2 in for an exponent.. so, 16 of an exponent of 1/2.. it would be letter C. 4

6 0

3 years ago

Read 2 more answers

Someone plz help me :(

LiRa [457]

Hey there ;-)

It will be -13,

as the other numbers can also be there in the inner circle.

By Benjemin ☺️

6 0

3 years ago

In a sample of 42 burritos, we found a sample mean 1.4 lb and assumed that sigma equals.5. In a test of the hypothesis H subscri

Sergio [31]

Answer:

D. 0.9953 (Probability of a Type II error), 0.0047 (Power of the test)

Step-by-step explanation:

Let's first remember that a Type II error is to NOT reject H0 when it is false, and the probability of that occurring is known as β. On the other hand, power refers to the probability of rejecting H0 when it is false, so it can be calculated as 1 - β.

To resolve this we are going to use the Z-statistic:

Z = (X¯ - μ0) / (σ/√n)

where μ0 = 1.2

σ = 5

n = 42

As we can see in part A of the attached image, we have the normal distribution curve representation for this test, and because this is a two-tailed test, we split the significance level of α=0.01 evenly into the two tails, 0.005 in each tail, and if we look for the Z critical value for those values in a standard distribution Z table we will find that that value is 2.576.

Now we need to stablish the equation that will telll us for what values of X¯ will we reject H0.

Reject if:

Z ≤ -2.576 Z ≥ 2.576

We know the equation for the Z-statistic, so we can substitute like follows and resolve.

Reject if:

(X¯ - 1.2) / (5/√42) ≤ -2.576 (X¯ - 1.2) / (5/√42) ≥ 2.576

X¯ ≤ -0.79 X¯ ≥ 3.19

We have the information that the true population mean is 1.25, so we now for a fact that H0 is false, so with this we can calculate the probability of a Type II error: P(Do not reject H0 | μ=1.25)

As we can see in part B of the attached image, we can stablish that the type II error will represent the probability of the sample mean (X¯) falling between -0.79 and 3.19 when μ=1.25, and that represents the shaded area. So now we now that we are looking for P(-0.79 < X¯ < 3.19 | μ=1.25).

Because we know the equation of Z, we are going to standardize this as follows:

P ( (-0.79 - 1.25) / (5/√42) < Z < (3.19 - 1.25) / (5/√42) )

This equals to:

P(-2.64 < Z < 2.51)

If we go and look for the area under the curve for Z positive scores in a normal standart table (part C of attached image), we will find that that area is 0.9940, which represents the probability of a Type II error.

Therefore, the power of the test will be 1-0.9940 = 0.006

If we look at the options of answers we have, there is no option that looks like this results, which means there was a probable redaction error, so we are going to stay with the closest option to these values which is option D.

3 0

3 years ago