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pickupchik [31]
3 years ago
14

Simplify the expression.

Mathematics
1 answer:
lianna [129]3 years ago
7 0

Answer:

C

Step-by-step explanation:

because it is. i KNOw for sure its right though

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ANSWER PLEASEEE
PtichkaEL [24]

Step-by-step explanation:

-3x+2=11

-3x=11-2

-3x=9

x=9÷(-3)

5 0
3 years ago
Your class earned $120.00 Saturday afternoon by washing cars to raise money for the class trip. This is 1/4 of the money needed
Ilya [14]

Answer:

$360.00

(I hope this was helpful)

5 0
3 years ago
What is the equation of a circle with a center at (–4, 0) that passes through the point (–2, 1)? x2 + (y + 4)² = StartRoot 5 End
Allisa [31]

Answer:

10

Step-by-step explanation:

no

6 0
4 years ago
Read 2 more answers
3x+y=1/3<br> 2x-3y=8/3<br><br> Solve the system of linear equation using any method.
krek1111 [17]
3x + 1y = ¹/₃   ⇒ 9x + 3y = 1
2x -  3y = 2²/₃ ⇒ 2x -  3y = 2²/₃
                                 11x = 3²/₃
                                  11     11
                                     x = ¹/₃
                             3x + y = ¹/₃
                         3(¹/₃) + y = ¹/₃
                               1 + y = ¹/₃
                             - 1        - 1
                                     y = ⁻²/₃
                              (x, y) = (¹/₃, ⁻²/₃)
4 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
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