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2 years ago

14

A spinner is divided into five equal sections numbered 1 through 5. Predict how many times out of 240 spins the spinner is most

likely to stop on an odd number.

1 answer:

pashok25 [27]2 years ago

3 0

If its has to be odd it would most likely be 29 times instead of 28

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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

Bodhi has a collection of 175 dimes and nickels. The collection is worth $13.30. Which equation can be used to find n, the numbe

Ede4ka [16]

Answer:

N + D = 175

.05N + .10D = $13.30

Step-by-step explanation:

You need a system of equations to get the correct answer that applies to both constraints.

6 0

2 years ago

Read 2 more answers

1) A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2,

crimeas [40]

Answer:

Step-by-step explanation:

Let us record the station number 1, 2 or 3 for each family member A, B or C.

I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.

From table it is clear that the total outcomes possible are 27.

We know that, probability can be defined as,

$PROBABITILY = \frac{NUMBER\;OF\;DESIRED\;OUTCOMES}{TOTAL\;NUMBER\;OF\;OUTCOMES}$

a) All Members Assigned to the Same Station.

Cases for all members being assigned to same station are as follows:

[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).

Therefore,

$PROBABILITY\;(Case\;a) = \frac{3}{27}\\\\PROBABILITY\;(Case\;a) = 0.111$

b) At Most Two Members Assigned to the Same Station.

It means that maximum of 2 members can have the same station. Cases for this situation are as follows:

[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],

[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]

(outcome number 2, 3, 4, 5, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 25 and 26 in the table).

Therefore,

$PROBABILITY\;(Case\;b) = \frac{18}{27}\\\\PROBABILITY\;(Case\;b) = 0.666$

c) All Members Assigned to a Different Station.

For this scenario, we have the following results:

[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).

Therefore,

$PROBABILITY\;(Case\;c) = \frac{6}{27}\\\\PROBABILITY\;(Case\;c) = 0.222$

3 0

2 years ago

Jenny is either a hippie or an artist. But Jenny is certainly not a hippie. So it can be concluded that Jenny is an artist is th

Natasha2012 [34]

Answer:

Deductive

Step-by-step explanation:

Jenny is an artist.

This is a deductive argument because the inductive use the particular to get the general and the deductive use the general to get the particular.

4 0

2 years ago

For her vacation Mrs. Andrews bought $300 worth of traveler's checks in $10 and $20 denominations

Nookie1986 [14]

Option C

The expression 20(22 - x) represents the value of the $20 traveler's checks she has.

<u>Solution:</u>

Given, For her vacation Mrs. Andrews bought $300 worth of traveler's checks in $10 and $20 denominations travelers checks in all,

We have to find how many of each denomination does she have?

Let x represent the number of $10 traveler's checks she has,

Then, $10(x) + $20(number of $20 checks) = 300

10x + 20($20 checks) = 300

x + 2(number of $20 checks) = 30 ⇒ (1)

Now let us see the given options, for value of $20 checks,

It should be in format, that, $20 x number of $20 checks

⇒ 20(total checks - $10 checks) ⇒ 20(total checks – x), as we can see only C is in that format, it is correct and total number of checks = 22.

Then, (1) ⇒ x + 2(22 – x) = 30

⇒ x + 44 – 2x = 30

⇒2x – x = 44 – 30

⇒ x = 14

So, $10 checks count = 14, then, $20 checks count = 22 – 14 = 8.

Hence, there are 14 $10 checks and 8 $20 checks, and option c is correct.

7 0

2 years ago