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3 years ago

Help pleeeeeeeeeaaaaaaseee

Mathematics

1 answer:

Helen [10]3 years ago

8 0

Group A has a greater maximum and it is more consistent than Group B.
The IQR is 700 for group A
The IQR is 750 for group B

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Write an equation in slope-intercept form, (0,1),(2,4)

Lorico [155]

Answer: y= 3/2x+1

Step-by-step explanation:

1) look for the y-intercept: It's given here in one of the ordered pairs (0,1)

2) find the slope: y2-y1/x2-x1 4-1/2-0= 3/2

3) substitute the slope and intercept for m and b

5 0

3 years ago

7÷42.28 <img src="https://tex.z-dn.net/?f=7%20%5Csqrt%7B42.28%7D%20" id="TexFormula1" title="7 \sqrt{42.28} " alt="7 \sqrt{42

Ksju [112]

0.166 not 166 hope it helps plz mark as brainliest

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2 years ago

List all number sets that apply to each number

Tresset [83]

What’s the numbers/ list ?

8 0

3 years ago

When you have two shapes to compare on a coordinate plane, you can determine the scale factor, knowing that the transformation w

musickatia [10]

Assume that the initial coordinates are (x,y) and that the dilated coordinates are (x',y').

The dilation is therefore:
(x,y) ............> (x',y')

Now, let's assume that the dilation factor is k.
Therefore:
x' = kx
y' = ky

Based on the above, all the student has to do is get the initial coordinates and the final ones and then substitute in any of the above two equations to get the value of k.

Example:
Assume an original point at (2,4) is dilated to coordinates (4,8). Find the dilation factor.
Assume the dilation coefficient is k.
(x,y) are (2,4) and (x',y') are (4,8)
Therefore:
x' = kx .........> 4 = k*2 ..........> k = 2
or:
y' = ky ..........> 8 = k*4 .........> k = 2
Based on the above, the dilation coefficient would be 2.

Hope this helps :)

8 0

3 years ago

Read 2 more answers

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let $\mu$ = true average percentage of organic matter

So, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is something other than 3%}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

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

where, $\bar X$ = sample mean percentage of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, the test statistics = $\frac{2.481-3}{\frac{1.616}{\sqrt{30} } }$ ~ $t_2_9$

= -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

8 0

3 years ago