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ruslelena [56]
3 years ago
15

Which graph is a histogram

Mathematics
1 answer:
Naya [18.7K]3 years ago
7 0

Answer:

The last one is a histogram. Histograms usually have that bar shape, and shows the fluctuation over time.

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What is a 15 yard gain 
Alex_Xolod [135]
Are you by any chance referring to football? If you are, it is either a play in which by running or passing the offense moves down the field towards the opponent's endzone 15 yards, or it can be obtained by the defense committing an egregious penalty, such as a facemask, unsportsmanlike conduct, or unnecessary roughness.
7 0
3 years ago
Suppose that Y varies directly with X andY equals 12 when X equals -2 what is X when Y equals -6
VLD [36.1K]

Answer:

x=1

Step-by-step explanation:

A direct variation equation has the following format:

y=kx

Where k is a constant.

We know that y is 12 when x is -2. Thus, substitute:

12=-2k

Divide both sides by -2:

k=-6

So, our direct variation equations is:

y=-6x

To find what x is when y equals -6, substitute in -6 for y:

-6=-6x

Divide both sides by -6:

x=1

So, our answer is 1 :)

4 0
3 years ago
If B = 0 in a linear equation of the form Ax + By = C, then which statement is also true? A. The graph of this equation is a hor
Vika [28.1K]

Answer:

<h3><u>A) Its horizontal line </u></h3>

Step-by-step explanation:

Ax + By = C

so Ax + (0)y = C

a= c /x

its horizontal

5 0
2 years ago
Pls help :)<br> What x what x what = 9 2/5
anyanavicka [17]

Answer:

1.5326

Step-by-step explanation:

there is no more simplification it can only be calculated by a calculator

8 0
2 years ago
Read 2 more answers
Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treat
stiks02 [169]

Answer:

Part A

b. 14.6 ± 7.38

Part B

b. 3.43

Part C

a. P-value < 0.01

Part D

b. There is sufficient evidence to reject the null hypothesis

Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

The number of seedlings selected to receive herbicide A = 10

The number of seedlings selected to receive herbicide B = 10

The height in centimeters of seedlings treated with Herbicide A, \overline x _1 = 94.5 cm

The standard deviation, s₁ = 10 cm

The height in centimeters of seedlings treated with Herbicide B, \overline x _2 = 109.1 cm

The standard deviation, s₂ = 9 cm

The 90% confidence interval for μ₂ - μ₁, is given as follows;

\left (\bar{x}_{2}- \bar{x}_{1}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

t_{(0.025, \, 18)} = 1.734

C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

b. 14.6 ± 7.38

Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

Hₐ; μ₂ - μ₁ ≠ 0

The two sample t-statistic is given as follows;

t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}

t-statistic=\dfrac{(109.1-94.5)}{\sqrt{\dfrac{10^{2} }{10}+\dfrac{9^{2}}{10}}} \approx 3.43173361147

The two sample t-statistic ≈ 3.43

b. 3.43

Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.

4 0
2 years ago
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