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7nadin3 [17]
2 years ago
5

-4 4/5 is equal to -4.7

Mathematics
1 answer:
Paha777 [63]2 years ago
3 0

Answer: Greater  

Step-by-step explanation: the first number is  

greater than the second number

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Molodets [167]
I hope this helps you

5 0
3 years ago
Given the equation y = 3x − 4, what is the value of x when y = 5? A) 3 B) 6 C) 9 D) 12
timurjin [86]

Answer:

3

Step-by-step explanation:

the equation would be 5 = 3(3) - 4

which is 9 - 4 = 5

7 0
3 years ago
(solving for y) <br> 4x-5=7+4y
alexira [117]
y = x - 3

1. Subtract 7 from the right to the left.
* 4y = 4x - 12
2. Divide the 4 (constant) next to the 'y' to the other side.
* y = 4/4x - 12/4 
* y = x - 3

3 0
3 years ago
What is 14 divided by 1,050
yanalaym [24]
It is 0.0133
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75
4 0
3 years ago
Read 2 more answers
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
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