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2 years ago

Write an expression that shows how to multiply 7 x 256 using expanded form and Distributive property.

1 answer:

6 0

To multiply any number <span>using expanded form and Distributive property, we need two steps:
1. Write the greater factor in expanded form, and
2. Record partial products by using Distributive property.

Step 1.
</span><span>7 x 256 = 7 x (200 + 50 + 6)
</span>
Step 2.
7 x (200 + 50 + 6) = 7 x 200 + 7 x 50 + 7 x 6

Hence, the complete expression should look like:
7 x 256 = 7 x (200 + 50 + 6) = 7 x 200 + 7 x 50 + 7 x 6

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J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0

2 years ago

Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The

abruzzese [7]

The question is incorrect.

The correct question is:

Three TAs are grading a final exam.

There are a total of 60 exams to grade.

(c) Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The TAs grade at different rates, so the first TA will grade 25 exams, the second TA will grade 20 exams and the third TA will grade 15 exams. How many ways are there to distribute the exams?

Answer: 60!/(25!20!15!)

Step-by-step explanation:

The number of ways of arranging n unlike objects in a line is n! that is ‘n factorial’

n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1

The number of ways of arranging n objects where p of one type are alike, q of a second type are alike, r of a third type are alike is given as:

n!/p! q! r!

Therefore,

The answer is 60!/25!20!15!

6 0

2 years ago

Jackson goes to the gym 0, 2, or 3 days per week, depending on work demands. The expected value of the number of days per week t

valina [46]

For a probability distribution the expected value is the summation of product of probabilities with their respective data values. Let x be the probability that Jackson goes gym for 2 days and y be the probability that he goes gym for 3 days.

For the given case we have following values and their probabilities:

0 : 0.1

2 : x

3 : y

So the expected value will be = 0(0.1) + 2(x) + 3(y)

Expected value is given to be 2.05. So we can write the equation as:

2x + 3y = 2.05 (Equation 1)

Also for a probability distribution, the sum of probabilities must always equal to 1. So we can set up the second equation as:

0.1 + x + y = 1

x + y = 0.9 (Equation 2)

From Equation 2 we can write the value of x to be x = 0.9 - y. Using this value in equation 1, we get:

2(0.9 - y) + 3y = 2.05

1.8 - 2y + 3y = 2.05

1.8 + y = 2.05

y = 0.25

Using the value of y in equation 2 we get value of x to be 0.65

Therefore we can conclude that:

The probability that Jackson goes to gym for 2 days is 0.65 and the probability that he goes to gym for 3 days is 0.25

6 0

2 years ago

Read 2 more answers

Find the 89th term of 25,35,45

MAXImum [283]

Answer:

The 89th term would be 895.

Step-by-step explanation:

There is a pattern of adding 10 each time. Notice that the 5 at the end doesn't change and that the first number continues (i.e. 2, 3, 4, 5, 6). All you have to do is add an 89 in front of the 5 and that is your answer, 895.

4 0

2 years ago

Read 2 more answers

4. In the following diagram, STU, RTP and

Papessa [141]

Yea I don’t know sorry bye

6 0

2 years ago