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alexdok [17]
3 years ago
14

A bakery made 140 buns. A restaurant bought 80 of the buns. The remaining buns were put into bags of 4 buns each. How many bags

of buns
are there?
Mathematics
2 answers:
Sedbober [7]3 years ago
7 0

Answer:

A bakery made 140 buns. A restaurant bought 80 of the buns. The remaining buns were put into bags of 4 buns each. How many bags of buns

are there?

Sloan [31]3 years ago
6 0

Answer:

30, 30 bags of buns your welcome

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Group of 7 people spend 33.50 movie tickets 6.50 adults and 3.50 for child ticket. Which system of equation could be used to fin
kramer
The answer is
x + y = 7
6.50x + 3.50y = 33.50

x - the number of adult tickets
y - the number of child tickets

<span>Group of 7 people: x + y = 7
</span> 6.50 adults and 3.50 for child ticket cost <span>33.50: 6.50x + 3.50y = 33.50

So, we have two equations:
</span>x + y = 7
6.50x + 3.50y = 33.50
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Step-by-step explanation:

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2 years ago
Evaluate z + 2+ z for x = 2, y=-3, z = -4.
Artyom0805 [142]

Answer:

-6

Step-by-step explanation:

Are you sure that z + 2+ z is correct?  x and y do not appear here.

z + 2+ z simplifies to 2z + 2, and so, if z = -4, z + 2+ z has the value

2(-4) + 2, or -6.

7 0
3 years ago
Please help me please hurry
Naya [18.7K]

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899

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8 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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