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3 years ago

13

At SaveRite, a 4-ounce carton of juice costs $0.85 and a 16-ounce carton costs $2.70. Which size carton would you buy to pay the

least amount per ounce?

1 answer:

Pie3 years ago

8 0

Answer:

16-ounce carton

Step-by-step explanation:

you divide by the cost by the amount of ounces per cent and when buying the 4 ounce one is 21 cents per ounce while the 16 ounce one is 16 cents per ounce

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Help !!!<br>See question in image.<br>Please show workings .<br>

ivolga24 [154]

Answer:

see explanation

Step-by-step explanation:

Given f(x) then the derivative f'(x) is

f'(x) = lim(h tends to 0 ) $\frac{f(x+h)-f(x)}{h}$

= lim ( h to 0 ) $\frac{4(x+h)^2-2-(4x^2-2)}{h}$

= lim ( h to 0 ) $\frac{4(x+h)^2-2-4x^2+2}{h}$

= lim( h to 0 ) $\frac{4x^2+8hx+4h^2-4x^2}{h}$

= lim( h to 0 ) $\frac{8hx+4h^2}{h}$

= lim ( h to 0 ) $\frac{4h(2x+h)}{h}$ ← cancel h on numerator/ denominator

= lim ( h to 0 ) 4(2x + h) ← let h go to zero

f'(x) = 8x

5 0

2 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$

= $\frac{3!}{1! \times 2!}\times \frac{2!}{1! \times 1!} \times \frac{2! \times 45!}{47!}$

= $\frac{6}{1081}$ = 0.0055

Now, probability that he ends up with a full house = $\frac{3}{1081} + \frac{6}{1081}$

= $\frac{9}{1081}$ = <u>0.0083</u>.

3 0

3 years ago

Read 2 more answers

From the moment the first car started how many hours will it take the second car to catch up to the first

Rina8888 [55]

Answer:

HELP

Step-by-step explanation:

3 0

3 years ago

Please answer this relatively quickly, I'm not in very much of a rush but would like to get this done.

aksik [14]

The answers is X = 23

8 0

2 years ago

Read 2 more answers

A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statisti

jeyben [28]

Answer:

We need a sample of size at least 13.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

90% confidence interval: (0.438, 0.642).

The proportion estimate is the halfway point of these two bounds. So

$\pi = \frac{0.438 + 0.642}{2} = 0.54$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?

We need a sample of size at least n.

n is found when M = 0.08. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.08 = 1.96\sqrt{\frac{0.54*0.46}{n}}$

$0.08\sqrt{n} = 1.96\sqrt{0.54*0.46}$

$\sqrt{n} = \frac{1.96\sqrt{0.54*0.46}}{0.08}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.54*0.46}}{0.08})^{2}$

$n = 12.21$

Rounding up

We need a sample of size at least 13.

3 0

3 years ago