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Alex Ar [27]
4 years ago
9

Why do banks charge fees?

Mathematics
1 answer:
Ivahew [28]4 years ago
3 0
They charge fees so they can cover there costs and return value to shareholders
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Answer

Step-by-step explanation:

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3 years ago
All true statements for the expression 4x+9-y(6x+7)
Sladkaya [172]

Answer: -6xy -7y +4x +9

Step-by-step explanation:

Firstly, distribute y to 6x + 7: -6xy - 7y

Next, collect like terms: 4x + 9 -6xy - 7y = -6xy -7y + 4x + 9

8 0
3 years ago
Jack is using a ladder to hang lights up in his house.he places the ladder 5 feet from the base of his house and leans it so it
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You did not complete your question but I think this is what you are looking for. View the image.

5 0
3 years ago
What is the value of x?
Paul [167]

Answer:

Step-by-step explanation:

5 0
4 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
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