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lara [203]
3 years ago
5

A container of oregano is 17 pounds heavier than a container of peppercorns. Their total weight is 253 pounds. The peppercorns w

ill be sold in one-ounce bags. How many bags of peppercorns can be made?
Mathematics
1 answer:
Ksju [112]3 years ago
5 0

Answer:

1,888 bags.

Step-by-step explanation:

Let x represent the the weight of container of peppercorns.

We have been given that the container of oregano is 17 pounds heavier than a container of peppercorns. So the weight of container of oregano would be x+17.

We are also told that the total weight of both containers is 253 pounds. We can represent this information in an equation as:

x+x+17=253

Let us solve for x.

2x+17=253

2x+17-17=253-17

2x=236

\frac{2x}{2}=\frac{236}{2}

x=118

Therefore, the weight of peppercorns is 118 pounds.

Now, we will convert 118 pounds to ounces.

1 pound equals 16 ounces.

118 pounds = 118*16 ounces = 1,888 ounces

Since the peppercorns will be sold in one-ounce bags, therefore, 1,888 bags of peppercorns can be made.

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3 years ago
The bakery used 12 cups of sugar in their large batch of cupcakes. How many quarts of sugar is this?
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3

Step-by-step explanation:

There are 4 cups in 1 quart, so 12/4 is just 3

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Step-by-step explanation:

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3 years ago
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The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

\hat p=\frac{X}{n}=\frac{224}{269}=0.8327

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

1-\hat p=1-0.8327

        =0.1673

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

\hat q=1-\hat p

  =1-\frac{X}{n}

  =1-\frac{546}{709}

  =0.2299\\\approx 0.229

The critical value of <em>z</em> for 95% confidence interval is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the 95% confidence interval for the population proportion as follows:

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

     =0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

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3 years ago
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Answer:5

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