Can anyone please help me? please?

Darius is a graphic designer he makes a sign for a company that has a width of 2 feet and a length of 2 feet 6 inches the compan

Sav [38]

Answer:

$\frac{1}{3}$

Step-by-step explanation:

Width of the sign made by Darius initially = 2 feet

Length of the sign = 2 feet 6 inches

Dimension of the flyer are 10 inches by 8 inches. Since, in original sign the measure of length is greater, therefore, the length of flyer is 10 inches and its width is 8 inches.

In order to find the scale factor we must convert the lengths and widths to same units. Lets convert the length and width of sign into inches.

Since, 1 feet = 12 inches

Length of the sign = 2 feet 6 inches = 2(12) + 6 inches = 30 inches

Width of the sign = 2 feet = 2(12) inches = 24 inches

Now we can find the scale factor by either comparing the lengths or widths of both the designs. Scale factor will be equal to the ratio of corresponding lengths/widths.

So, the scale factor would be:

Length of Flyer : Length of Sign

= 10 inches : 30 inches

= 1 : 3

= $\frac{1}{3}$

This shows, the length of flyer is $\frac{1}{3}$ times as that of the sign. So, the scale factor that Darius must use is $\frac{1}{3}$ . The length and width of the flyer are $\frac{1}{3}$ as that of the sign.

The same scale factor would result if we would have used the ratio of widths instead of the lengths.

Scale Factor = Width of Flyer : Width of sign

= 8 : 24

= 1 : 3

Therefore, Darius must type the scale factor of $\frac{1}{3}$ in his computer to get the size of the flyer.

7 0

2 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.