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oksano4ka [1.4K]

3 years ago

13

If 4 items are chosen at random without replacement from 7 items, in how many ways can the 4 items be arranged, treating each ar

rangement as a different event (i.e., if order is important)?
A. 35
B. 840
C. 5040
D. 24

1 answer:

vodka [1.7K]3 years ago

8 0

Answer:

840

Step-by-step explanation:

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You design a new app for cell phones. Your revenue for x downloads is given by f(x) = 3x Your profit

Snowcat [4.5K]

Answer:

p = 2.7x - 20

p(90) = $223

Step-by-step explanation:

revenue → f(x) = 3x ,where x is the number of downloads.

the statement “Your profit p is $20 less than 90% of the revenue for x downloads”

<u><em>means</em></u>

$p=90\% f\left( x\right) -20$

<u><em>means</em></u>

<u><em></em></u> $p=\frac{90}{100} \times f\left( x\right) -20$

<u><em>means</em></u>

$p=\frac{9}{10} \times (3x) -20$

<u><em>means</em></u>

$p=\frac{9}{10} \times (3x) -20$

<u><em>means</em></u>

$p=2.7x -20$

………………………………

the profit for 90 downloads :

= p(90)

= 2.7×(90) - 20

= $223

5 0

2 years ago

A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true prop

attashe74 [19]

Answer:

$0.424 - 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.286$

$0.424 + 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.562$

The 99% confidence interval would be given by (0.286;0.562)

Step-by-step explanation:

Information given:

$X= 36$ represent the families owned at least one DVD player

$n= 85$ represent the total number of families

$\hat p=\frac{36}{85}= 0.424$ represent the estimated proportion of families owned at least one DVD player

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.424 - 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.286$

$0.424 + 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.562$

The 99% confidence interval would be given by (0.286;0.562)

8 0

3 years ago

A standard deck of playing cards has 52 cards: 13 spades, 13 clubs, 13 hearts, and 13 diamonds. What is the probability of drawi

cestrela7 [59]

13/ 52 = 1/4 chances.

There are 13 spades and 52 cards. You divide the spades and cards in the deck to find your chances of drawing one.

You have a 25% chance of getting a spade.

Now you put it back. Since you put it back, you still have the same number of cards and the same number of spades. So when you divide 13/52 again, the numbers are still the same. 1/4. 25% chance.

So, you still have a 25% chance the second time around.

5 0

4 years ago

Help please 50 PTS! I need this asap

mars1129 [50]

Answer:

How tall are you

ez clappp

6 0

2 years ago

Cindy won the lottery, and she gave her family $1.000.00 less than of her winnings. If the jackpot was worth $35.000.00, how muc

Leviafan [203]

That would be $34.000.000 since it’s just subtracting ( less than ) the main number minus the number she didn’t give away.

7 0

3 years ago