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3 years ago

To make 5 apple pies, you need about 2 pounds of apples. How many pounds of apples do you need to make 20 apple pies?

Mathematics

2 answers:

Lyrx [107]3 years ago

7 0

About 8 pounds of apples is needed for 20 pies
<span> 20 divided by five is 4 so you have to multiply 2 times 4 </span>

nlexa [21]3 years ago

5 0

8 lbs. We have 2 lbs for every 5 pies so the first step is to find out how many sets of 5 we have. We find this out by dividing 20 by 5. 20 divided by 5 is 4, so we have 4 sets of 5. If for every one set, we need 2 lbs of apples, we need to multiply the number of sets by 2. 4*2 is 8, so 8 is our final answer.

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Suppose that 16% of the population of the U.S. is left-handed. If a random sample of 170 people from the U.S. is chosen, approxi

Shalnov [3]

Answer:

0.6064 = 60.64% probability that fewer than 29 are left-handed.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

16% of the population of the U.S. is left-handed.

This means that $p = 0.16$

Sample of 170 people

This means that $n = 170$

Mean and standard deviation:

$\mu = E(X) = np = 170*0.16 = 27.2$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{170*0.16*0.84} = 4.78$

Probability that fewer than 29 are left-handed.

Using continuity correction, this is P(X < 29 - 0.5) = P(X < 28.5), which is the pvalue of Z when X = 28.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{28.5 - 27.2}{4.78}$

$Z = 0.27$

$Z = 0.27$ has a pvalue of 0.6064

0.6064 = 60.64% probability that fewer than 29 are left-handed.

6 0

3 years ago

Violet bought 7 air filters at the store. Each filter cost $18.85. Violet used a coupon for $5.50 off her total cost of the air

liubo4ka [24]

$126.45
7x18.85 - coupon

8 0

3 years ago

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Please help! I don’t get it much

Shkiper50 [21]

9514 1404 393

Answer:

∠BDC = 46°

Step-by-step explanation:

Angle BDC intercepts arc BC, identified as having a measure of 92°. The angle's vertex, point D, is on the circle, so the angle is called an "inscribed angle." The measure of an inscribed angle is half the measure of the arc it intercepts.

∠BDC = (1/2)arc BC = 1/2(92°)

∠BDC = 46°

3 0

3 years ago

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

3 years ago

10^-2÷(5×10^-3)<br>^ means power

lorasvet [3.4K]

Answer:

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3 years ago

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