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mezya [45]
3 years ago
8

Jake wanted to measure the width of the pond, so

Mathematics
1 answer:
mr_godi [17]3 years ago
5 0

Answer:

C

Step-by-step explanation:

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agasfer [191]

c = cost

h = # of half-dozens ordered

c = $8.50h + $1  is the correct answer.

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solve this math problem. Estimate it by giving whole number. Show working also. (ESTIMATE, NOT WORK OUT) :)​
pentagon [3]

Answer:

(a) 11.

(b) 5.

Step-by-step explanation:

(a)√124:

11*11 = 121 so an estimate is 11.

(b) ∛124

5*5*5 = 125

So an estimate is 5.

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3 years ago
What is the solution to the equation?
stiv31 [10]

Answer:

x = - 71

Step-by-step explanation:

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7 0
3 years ago
If 200g of fat is used for 8 cakes, how much is used for 3 cakes?
Mrrafil [7]
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3 years ago
Read 2 more answers
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
Viefleur [7K]

Answer:

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

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}.

For this problem, we have that:

Suppose a sample of 1537 tenth graders is drawn. Of the students sampled, 1184 read above the eighth grade level. So 1537 - 1184 = 353 read at or below this level. Then

n = 1537, \pi = \frac{353}{1537} = 0.2297

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.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 - 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2087

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 + 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2507

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

7 0
3 years ago
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