Mary wrote a book of 1536 pages. She numbered all pages by hand. How many times did she write the digit 6?

A department store purchases logo shirts at a cost of $10 per shirt. They increase the price 100% and put them on the sales floo

Andrei [34K]

The original selling price is found by multiplying 100% by $10 and adding that to the original $10. The logo shirt goes on the sale floor for $20. Each of the next three months, the reduced price is found by multiplying the current sale price by 75%. The tax is found by multiplying 5% by the final discounted price and adding the tax to the final sale price.

7 0

2 years ago

Read 2 more answers

Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The

Harlamova29_29 [7]

Answer:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

We have the following data given:

$n_1 =20$ represent the sample size for group 1

$n_2 =20$ represent the sample size for group 2

$\bar X_1 =43.5$ represent the sample mean for the group 1

$\bar X_2 =40.1$ represent the sample mean for the group 2

$s_1=4.1$ represent the sample standard deviation for group 1

$s_2=3.2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525$

And the deviation would be just the square root of the variance:

$S_p=3.678$

The statistic is givne by:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0

2 years ago

Anumeha is mowing lawns for a summer job. For every mowing job, she charges an initial fee of $ 10 $10dollar sign, 10 plus a con

slavikrds [6]

Answer:

F(t) = 10 + 5(t)

Step-by-step explanation:

The complete question is as follows;

Anumeha is mowing lawns for a summer job. for every mowing job, she charges an initial fee of \$10$10dollar sign, 10 plus a constant fee for each hour of work. her fee for a 555-hour job, for instance, is \$35$35dollar sign, 35. let f(t)f(t)f, left parenthesis, t, right parenthesis denote anumeha's fee for a single job fff (measured in dollars) as a function of the number of hours ttt it took her to complete it. write the function's formula.

Solution

We are interested in writing the function F(t) formula for the fee charged by Anumeha per job.

Now, the key to writing this function is knowing exactly the constant fee she charges on the job.

We were told that she got $35 for a 5 hour job.

Thus, the constant amount charged is as follows;

Since it’s $10 as initial fee and the constant fee is per hour;

35 = 10 + 5(x)

where x is the constant fee per hour

35 = 10 + 5x

5x = 35-10

5x = 25

x = 25/5

x = $5

This mess that she charges a constant fee of $5 per hour

So we can write the equation now.

F(t) = 10 + 5(t)

where t represents the number of hours she spent on the job

6 0

2 years ago

RANDOMLY CHOOSING TWO POTENTIAL CLIENTS

gregori [183]

Let Jacob, Carol, Geraldo, Meg, Earvin, Dora, Adam, and Sally be represented by the letters J, C, G, M, E, D, A, and S respectively. 

In part IV we are asked:

What is the sample space of the pairs of potential clients that could be chosen?

Since the Sample Space is the set of all possible outcomes, we need to make a set (a list) of all the possible pairs, which are as follows:

{(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S)

, (C, G), (C, M), (C, E), (C, D), (C, A), (C, S)

 , (G, M), (G, E), (G, D), (G, A), (G, S)

 ,(M, E), (M, D), (M, A), (M, S)

, (E, D), (E, A), (E, S) 

, (D, A), (D, S)

, (A, S).}

We can check that the number of the elements of the sample space, n(S) is

1+2+3+4+5+6+7=28.

This gives us the answer to the first question: How many pairs of potential clients can be randomly chosen from the pool of eight candidates?

(Answer: 28.)

II) What is the probability of any particular pair being chosen?

The probability of a particular pair to be picked is 1/28, as there is only one way of choosing a particular pair, out of 28 possible pairs.

III) What is the probability that the pair chosen is Jacob and Meg or Geraldo and Sally?

The probability of choosing (J, M) or (G, S) is 2 out of 28, that is 1/14.

Answers:

I) 28
II) 1/28≈0.0357
III) 1/14≈0.0714
IV)

{(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S)
, (C, G), (C, M), (C, E), (C, D), (C, A), (C, S)
, (G, M), (G, E), (G, D), (G, A), (G, S)
,(M, E), (M, D), (M, A), (M, S)
, (E, D), (E, A), (E, S)
 , (D, A), (D, S)
, (A, S).}

6 0

2 years ago

When plucked, the high E string on a guitar has a frequency of 330 cycles per second. What sine function represents this note wh

Ber [7]

Let's analyse the function

$y = f(x) = A\sin(\omega x)$