Evaluate the expression x - 7. If x= 12. 19 6 17 5

Use a table to identify three equivalent ratios of 2/5.

zlopas [31]

Answer: The simplest form of ratio 6:15=2×3:5×3=2:5

Step-by-step explanation: So, 2:5 is equivalent to 6:15.

7 0

2 years ago

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A farmer applied different types of fertilizer to two different sections of his watermelon field. At the end of the growing seas

Softa [21]

The mean is the average weight.
the MAD is how spread out the weights are from the mean, the higher the number the more spread out, which means you could have more lower weights than higher weights, of more heavy than light

so fertilizer 1 had a higher average weight ( 24 is greater than 22)
and the difference in MAD is only 1 ( 3-2)

so the farmer CAN conclude , because the distribution is SOMEWHAT SIMILAR

3 0

4 years ago

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A train left the station. It traveled 30 miles and dropped off passengers. Then it turned around and traveled 15 miles. Write an

viva [34]

15 miles because it travelled 30 miles then travelled 15 miles in the direction of the station, so 30-15=15

6 0

3 years ago

The total weight of the fish in a tank of tropical fish at Fish’n’fur was 7/8 Ib. Each fish weighed 1/64 Ib. After Eric bought s

Sonbull [250]

Step-by-step explanation:

Weight of fish Eric bought

= Weight of fish at first - Weight remaining

= 7/8 lb - 1/2 lb = 3/8 lb.

Number of fish Eric bought

= (3/8 lb) / (1/64 lb)

= (24/64 lb) / (1/64 lb)

= 24.

Hence Eric bought 24 fishes.

5 0

3 years ago

A presidential candidate's aide estimates that, among all college students, the proportion who intend to vote in the upcoming el

inessss [21]

Answer:

$z=\frac{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24$

$p_v =P(z$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6

Step-by-step explanation:

Assuming the following question: A presidential candidate's aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at least 60% . If 127 out of a random sample of 240 college students expressed an intent to vote, can we reject the aide's estimate at the 0.1 level of significance?

Data given and notation

n=240 represent the random sample taken

X=127 represent the college students expressed an intent to vote

$\hat p=\frac{127}{240}=0.529$ estimated proportion of college students expressed an intent to vote

$p_o=0.6$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that at least 60% of students are intented to vote .:

Null hypothesis: $p \geq 0.6$

Alternative hypothesis: $p < 0.6$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6

7 0

3 years ago