Grace has 1.35 pounds of strawberries, 1.4 pounds of bananas, and some apples. She has more pounds of apples than pounds of stra
1 answer:
Answer:
Grace could have 1.36 pounds, 1.37 pounds, 1.38 pounds or 1.39 pounds of apples.
Step-by-step explanation:
Let a represent pounds of apples.
We have been given that Grace has 1.35 pounds of strawberries. She has more pounds of apples than pounds of strawberries.
This means that a is greater than 1.35. We can represent this information in an inequality as:
We are also told that Grace has 1.4 pounds of bananas. She has fewer pounds of apples than pounds of bananas. This means that a is less than 1.4. We can represent this information in an inequality as:
Upon combining both inequalities, we will get:
Therefore, Grace could have 1.36 pounds, 1.37 pounds, 1.38 pounds or 1.39 pounds of apples.
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Answer:
We conclude that a web-based company are not exceeding their goal of 90%.
Step-by-step explanation:
We are given that a web-based company has a goal of processing 90 percent of its orders on the same day they are received.
434 out of the next 471 orders are processed on the same day.
Let p = <u><em>proportion of orders processing on the same day they are received.</em></u>
SO, Null Hypothesis, : p
0.90 {means that they are not exceeding their goal of 90%}
Alternate Hypothesis, : p > 0.90 {means that they are exceeding their goal of 90%}
The test statistics that would be used here <u>One-sample z test for</u> <u>proportions</u>;
T.S. = ~ N(0,1)
where, = sample proportion of orders that are processed on the same day =
= 0.92
n = sample of orders = 471
So, <u><em>the test statistics</em></u> =
= 1.599
The value of z test statistics is 1.599.
<u>Now, at 0.025 significance level the z table gives critical value of 1.96 for right-tailed test.</u>
Since our test statistic is less than the critical value of z as 1.599 < 1.96, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.
Therefore, we conclude that a web-based company are not exceeding their goal of 90%.
They could have assembled total 336 different possible crews.
<em><u>Explanation</u></em>
Total number of trained pilots is 9 and geologists is 4.
The crew should be consisted of 2 pilots and 1 geologist. That means, <u>2 pilots are chosen from total 9 and 1 geologist is chosen from total 4</u>.
Number of ways for choosing 2 pilots from 9 and
Number of ways for choosing 1 geologist from 4
So, the total possible number of different crews they could have assembled