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1 year ago

A wagon can hold 44 pounds. A newspaper weighs 22 ounces. How many newspapers can the wagon hold?

Mathematics

2 answers:

miss Akunina [59]1 year ago

6 0

Answer:

The wagon can hold 32 newspapers

Step-by-step explanation:

Digiron [165]1 year ago

5 0

The wagon can hold 32 newspapers. There are 16 ounces in a pound so multiply 44*16 and then divide it by 22

You might be interested in

Alice and Bob each have a certain amount of money. If Alice receives $n$ dollars from Bob, then she will have $4$ times as much

oksano4ka [1.4K]

Answer: 31 : 9

Step-by-step explanation:

Assume the following:

Alice's amount = P

Bob's amount = Q

Amount received = n

If Alice receives $n$ dollars from Bob ;then she will have $4$ times as much money as Bob.

P + n = 4(Q - n)

P + n = 4Q - 4n

P = 4Q - 4n - n

P = 4Q - 5n - - - - (1)

If, on the other hand, she gives $n$ dollars to Bob, then she will have $3$ times as much money as Bob

P - n = 3(Q + n)

P - n = 3Q + 3n

P = 3Q + 3n + n

P = 3Q + 4n - - - - - - (2)

Equating both equations - (1) and (2)

4Q - 5n = 3Q + 4n

4Q - 3Q = 4n + 5n

Q = 9n

Express P in terms of n, use either equation (1) or (2)

From equation 2:

P = 3Q + 4n

Substituting Q = 9n

P = 3(9n) + 4n

P = 27n + 4n

P = 31n

Alice's amount = P, Bob's = Q

Ratio = P:Q

31 : 9

5 0

2 years ago

A Washington, D.C., "think tank" announces the typical teenager sent 67 text messages per day in 2017. To update that estimate,

creativ13 [48]

Answer:

We reject the null hypothesis and fail to accept it and update that estimate that typical teenager sent does not 67 text messages per day.

Step-by-step explanation:

We are given the sample:

51, 175, 47, 49, 44, 54, 145, 203, 21, 59, 42, 100

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{990}{12} = 82.5$

Sum of squares of differences = 992.25 + 8556.25 + 1260.25 + 1122.25 + 1482.25 + 812.25 + 3906.25 + 14520.25 + 3782.25 + 552.25 + 1640.25 + 306.25 = 3539.363636

$S.D = \sqrt{\frac{3539.363636}{11}} = 59.5$

We are given the following in the question:

Population mean, μ =67

Sample mean, $\bar{x}$ = 82.5

Sample size, n = 12

Alpha, α = 0.05

Sample standard deviation, s = 59.5

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 67\\H_A: \mu > 67$

We use One-tailed t test to perform this hypothesis.

b) Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n-1}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{82.5 - 67}{\frac{59.5}{\sqrt{11}} } = 0.864$ Now,

$t_{critical} \text{ at 0.05 level of significance, 11 degree of freedom } = 1.795$

a) Since,

$t_{stat} < t_{critical}$

We reject the null hypothesis and fail to accept it and update that estimate that typical teenager send more than 67 text messages per day.

5 0

1 year ago

A carload of steel rods has arrived at Cybermatic Construction Company. The car contains 50,000 rods. Claude Ong, manager of Qua

amid [387]

Answer:

There is a 99.24% probability that Claude's sample has a mean between 119.985 and 120.0125 inches.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples 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.

In this problem, we have that:

The population of rods has a mean length of 120 inches and a standard deviation of 0.05 inch. This means that $\mu = 120, \sigma = 0.05$ .

Claude Ong, manager of Quality Assurance, directs his crew measure the lengths of 100 randomly selected rods. This means that $n = 100, s = \frac{\sigma}{\sqrt{n}} = \frac{0.05}{\sqrt{100}} = 0.005$ .

The probability that Claude's sample has a mean between 119.985 and 120.0125 inches is

We are working with the sample mean, so we use the standard deviation of the sample, that is, $s$ instead of $\sigma$ in the z score formula.

This probability is the pvalue of Z when $X = 120.0125$ subtracted by the pvalue of Z when $X = 119.985$ .

X = 120.0125

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

$Z = \frac{120.0125 - 120}{0.005}$

$Z = 2.5$

$Z = 2.5$ has a pvalue of 0.9938.

X = 119.985

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

$Z = \frac{119.985 - 120}{0.005}$

$Z = -3$

$Z = -3$ has a pvalue of 0.0014.

So there is a 0.9938 - 0.0014 = 0.9924 = 99.24% probability that Claude's sample has a mean between 119.985 and 120.0125 inches.

7 0

2 years ago

There are 950 students at Hanover High School. The ratio of the number of freshmen to all students is 3:10. The ratio of the num

ss7ja [257]

Answer:

Step-by-step explanation:

There are 950 students at Hanover High School. The ratio of the number of freshmen to all students is 3:10. This means that the number of freshmen is 3/10 × total number of students.

Number of freshmen = 3/10×950 =285 freshmen

The ratio of the number of sophomore to all students is 1:2. This means that the number of sophomore is 1/2 × total number of students.

Number of sophomore = 1/2×950 =475 sophomore

The ratio of the number of freshmen to sophomore is the number of freshmen/ the number of

sophomore.

It becomes 285/475 = 57/95

6 0

1 year ago

You have been observing a population of swallowtail butterflies for the last 10 years. the size of the population has varied as

EastWind [94]

The answer to this questions is = 17

4 0

2 years ago