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kotegsom [21]
2 years ago
12

Write an expression that represents the number of shells in pile 2.

Mathematics
1 answer:
daser333 [38]2 years ago
5 0

Answer:

2

Step-by-step explanation:

4/2=2

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Osvoldo has a goal of getting at least 30% of his grams of carbohydrates each day from whole grains. Today, he ate 220 grams of
pochemuha

Answer:

Answer: Osvoldo did not meet his goal.

Step-by-step explanation:

Total carbohydrate ate by Osvoldo = 220 g

carbohydrate from grains = 55 g

Percentage of carbohydrate of grain is

\begin{gathered}=\dfrac{\text { carbohydrate taken from grains}}{\text {Total carbohydrate}} \times 100\\\\= \dfrac{55}{220} \times 100= 25 \%\end{gathered}

=

Total carbohydrate

carbohydrate taken from grains

×100

=

220

55

×100=25%

Goal of getting carbohydrate from grains = at least 30%

but he ate only 25% of carbohydrate

Hence, Osvoldo did not meet his goal as he ate less carbohydrate than his goal.

8 0
3 years ago
Barry asked 40 random students at his school to name their favorite ice cream flavor. Of the 40 students he asked, 25 of them pr
bagirrra123 [75]

Answer:

750

Step-by-step explanation:

62.5% choose chocolate

1200 x 62.5% of is 750

3 0
2 years ago
A telecommunication company has 7 satellites, two of which are sending a week signal. If two are picked at random without replac
Setler [38]
2/7 is the probability that the first satellite sends a weak signal. That leaves 6, one of which sends a weak signal. So the probability that the second also sends a weak signal is 1/6. Combine these and we get 2/7×1/6=1/21.
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
Hi could someone solve this for me i have been stuck for an hour now XD
joja [24]

Answer:

The anwser is 40.82

Step-by-step explanation:

Pie times radius

4 0
2 years ago
Read 2 more answers
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