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Lynna [10]
3 years ago
14

There are 30 students in Ms. Prema's history class. 18 of these students turned in an extra credit project at the end of the sem

ester.
Mathematics
1 answer:
kow [346]3 years ago
3 0

Answer:

12 students didnt turn in a extra credit project at the end of the semester.

Step-by-step explanation:

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Taya2010 [7]
Based on the given description above, here is how we are going to solve it.
Let n be the number of years. So the equation will be like this.
37000n + 1500n = 1025000
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38500n = 1025000 <<<divide by sides by 38500 and the result is
n=26.62
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8 0
3 years ago
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What conic section is drawn by the parametric equations x=csc t and y cot?
Vanyuwa [196]
If we take the square of x and square of y and then subtract them:
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 equation 'a' represent the equation of hyperbola which is (x²/a²)-(y²/b²) =1 with given conditions( a=1,b=1)

So, option D is correct
3 0
3 years ago
If your car used 19.2 gallons of gas to go 1293.6 miles, how many miles per gallon (mpg) did the car get?
Anna007 [38]
G=m/x
G-gallons
m-miles
x-miles per gallon
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x=1293.6/19.2
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5 0
3 years ago
An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so
alexgriva [62]

Solution :

Let p_1 and p_2  represents the proportions of the seeds which germinate among the seeds planted in the soil containing 3\% and 5\% mushroom compost by weight respectively.

To test the null hypothesis H_0: p_1=p_2 against the alternate hypothesis  H_1:p_1 \neq p_2 .

Let \hat p_1, \hat p_2 denotes the respective sample proportions and the n_1, n_2 represents the sample size respectively.

$\hat p_1 = \frac{74}{155} = 0.477419

n_1=155

$p_2=\frac{86}{155}=0.554839

n_2=155

The test statistic can be written as :

$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}

which under H_0  follows the standard normal distribution.

We reject H_0 at 0.05 level of significance, if the P-value or if |z_{obs}|>Z_{0.025}

Now, the value of the test statistics = -1.368928

The critical value = \pm 1.959964

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

                                     $=2 \times 0.085667$

                                     = 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$, so we fail to reject H_0 at 0.05 level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the \text{proportion} of the seeds that \text{germinate differs} with the percent of the \text{mushroom compost} in the soil.

8 0
3 years ago
Woodland Mountain Park sells annual visitor passes for 12:50 last year the Park Race 53 and 500 in an annual visiting sales how
Lorico [155]

Answer:

4,280 tickets were sold

Step-by-step explanation:

3 0
4 years ago
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