Based on the graph below, what is a reasonable estimate for the у value when x is 4?

The manufacturer of an airport baggage scanning machine claims it can handle an average of 553 bags per hour. (a-1) At α = .05 i

timurjin [86]

Answer:

H1 : μ < 553. Reject H1 if tcalc > –1.753.

There is not enough evidence to reject the manufacturer’s claim.

Step-by-step explanation:

We are given that the manufacturer of an airport baggage scanning machine claims it can handle an average of 553 bags per hour.

A sample of 16 randomly chosen hours with a mean of 533 and a standard deviation of 47 is given.

Let $\mu$ = average bags that an airport baggage scanning machine can handle.

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 553 bags {means that the manufacturer’s claim is not overstated}

Alternate Hypothesis, $H_0$ : $\mu$ < 553 bags {means that the manufacturer’s claim is overstated}

The test statistics that would be used here One-sample t-test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 533 bags

s = sample standard deviation = 47

n = sample of hours = 16

So, the test statistics = $\frac{533-553}{\frac{47}{\sqrt{16} } }$ ~ $t_1_5$

= -1.702

The value of t test statistics is -1.702.

Now, at 0.05 significance level the t table gives critical value of -1.753 at 15 degree of freedom for left-tailed test.

Since our test statistic is more than the critical value of t as -1.702 > -1.753, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the manufacturer’s claim is not overstated and an airport baggage scanning machine can handle an average of 553 bags per hour.

5 0

3 years ago

Use 3.14 for .Jenna is planting four circular gardens. She is planting one garden each for tomatoes, carrots, cabbage, and straw

Llana [10]

Um I assume you're wanting the circumferences of the fences so we have 3 that are all the same size and one that is 3 times as large as any of the others

so here's our equations

3x+y=113.04

y=3x

use substitution

3x+3x=113.04

6x= 113.04

now the question says use 3.14 so I'm thinking you want the answer in terms of pi so first divide 113.04 by 3.14

6x=36pi

x=6pi now substitute that back in for x in one of your equations

y=(3)(6pi)

y=18pi

so the tomatoes carrots and cabbage will all have a circumference of 6pi and the strawberries 18pi

7 0

3 years ago

A normal distribution has a mean of 15 and a standard deviation of 2. Find the value that corresponds to the 75th percentile. Ro

Sonbull [250]

Answer:

16.35

Step-by-step explanation:

Using an inverse normal distribution, we can calculate the z score given the percentile, which can then be used to find our value.

First, we can use an inverse normal distribution calculator to figure out that the z score given the 75th percentile is 0.674.

Next, we know that the z score is (observed value - mean) / standard deviation. We can plug our values in to get

$\frac{x-15}{2} = z\\\frac{x-15}{2} = 0.674\\x-15 = 0.674 * 2\\x = 0.674 * 2 + 15\\x= 16.348$

Rounding, we get x = 16.35 as our answer

5 0

3 years ago

God why can't I understand thus stuff

damaskus [11]

Hey there,
For this question, you need to find the lowest common multiple which in this case is 60.
How i got it?
5= 1 x 5
12= 2 square x 3
Since one is common, take 5 x 2 square x 3 which is equal to 60

Hope you get it :))

You can ask me any doubts if you have :))

~Top

7 0

3 years ago

If f(x)= x^2 which of the following describes the graphs of f(x) +4

deff fn [24]

The parabola was moved up 4 on the y-axis

7 0

3 years ago