3 and 5/6 minus 2 and 3/4

Rauol measured the height of 12 different plants in his garden in April. He measured them again in June. What is the difference

jasenka [17]

The question is incomplete. Here is the complete question.

Raoul measured the ehight of 12 different plants in his garden in April. He measured them again in June.

April Heights (in inches) 12 15 23 11 42 45 44 39 12 19 20 12

June Heights (in inches) 45 45 40 43 11 14 12 13 41 40 45 41

What is the difference between the mean height in April and the mean height in June?

Answer: The difference between means is 8.0

Step-by-step explanation: Mean is the average of a set of numbers.

It is calculated as:

$\mu=\frac{\Sigma x}{n}$

where

x is the numbers in the set

n is how many numbers are there in the set

For heights in April, the mean is:

$\mu_{1}=\frac{12+15+23+11+42+45+44+39+12+19+20+12}{12}$

μ₁ = 24.5

For heights in June, the mean os:

$\mu_{2}=\frac{45+45+40+43+11+14+12+13+41+40+45+41}{12}$

μ₂ = 32.5

The difference between means:

μ₂ - μ₁ = 32.5 - 24.5 = 8.0

The difference between the mean height in April and in June is 8.0 inches.

3 0

2 years ago

I need help with this question pic is above

zavuch27 [327]

A: peak
B: falling
C:on the ground

6 0

3 years ago

Which equation represents circle C?

OleMash [197]

Answer:

B

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Jennifer made 5 liter of punch for her for her party her brother made another 750ml if they combine the two batches how many 180

kykrilka [37]

5 liters = 5000ml

If they combine, they would have
5000 + 750 = 5750ml

one serving = 180ml
Number of serving = 5750 ÷ 180 = 31 serving with 170ml remaining

7 0

3 years ago

A professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their class

KIM [24]

Answer:

We conclude that seniors skip more than 2% of their classes at 0.01 level of significance.

Step-by-step explanation:

We are given that a professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their classes.

He randomly samples a group of seniors and out of 2521 classes, the group skipped 77.

Let p = percentage of seniors who skip their classes.

So, Null Hypothesis, $H_0$ : p $\leq$ 2% {means that seniors skip less than or equal to 2% of their classes}

Alternate Hypothesis, $H_A$ : p > 2% {means that seniors skip more than 2% of their classes}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of seniors who skipped their classes = $\frac{77}{2521}$

n = sample of classes = 2521

So, test statistics = $\frac{\frac{77}{2521} -0.02}{{\sqrt{\frac{\frac{77}{2521}(1-\frac{77}{2521})}{2521} } } } }$

= 3.08

The value of the test statistics is 3.08.

Now at 0.01 significance level, the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical value of z as 2.3263 < 3.08, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that seniors skip more than 2% of their classes.

6 0

3 years ago