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4 years ago

Someone help me asap pls

Mathematics

1 answer:

Makovka662 [10]4 years ago

8 0

Answer:

For the graph in the top-left corner, the ?? = 0.25

This can be found using a simple slope equation:

m = (y2-y1)/(x2-x1)

m = slope

You take two coordinates, in this case (8, 2) and (4, 1), label one as #1 and the other as #2, and plug in the values so:

slope = (2-1)/(8-4)

slope = 1/4 or 0.25

The equation of a line is y = mx + b and if you were to place 0.25, or 1/4, in place of m (the slope) and 1 in the place of x since it is the x value, and multiply, you would end up with 0.25.

This also gives you the answer to your next two questions.

The one in the top-right corner is $0.25 per pencil.

The answer to the question just below it is y = 0.25x.

b in the equation would equal 0 since b is the y-intercept (where the line crosses the y-axis)

As for the last one, you simply have to multiply 30 by each number in the left column since you already know 30 miles are travelled in 1 hour, giving you your rate or the equation of the line, y=x.

The missing values of the table are as follows:

0, 60, 150, 240, 330, 1470

Finally, the last box:

Miles = 30 * hours

y = 30x

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Suppose you are given either a fair dice or an unfair dice (6-sided). You have no basis for considering either dice more likely

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Answer: Our required probability is 0.83.

Step-by-step explanation:

Since we have given that

Number of dices = 2

Number of fair dice = 1

Probability of getting a fair dice P(E₁) = $\dfrac{1}{2}$

Number of unfair dice = 1

Probability of getting a unfair dice P(E₂) = $\dfrac{1}{2}$

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Probability of getting a 3 for the unfair dice P(A|E₂) = $\dfrac{1}{3}$

So, we need to find the probability that the die he rolled is fair given that the outcome is 3.

So, we will use "Bayes theorem":

$P(E_1|A)=\dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\\\\(E_1|A)=\dfrac{0.5\times 0.16}{0.5\times 0.16+0.5\times 0.34}\\\\P(E_1|A)=0.83$

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3 years ago

What is the area of triangle FGH? Round your answer to the nearest tenth of a square centimeter.

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3 years ago

5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g

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Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{1117}{10} = 111.7$

Sum of squares of differences = 642.1

$S.D = \sqrt{\frac{642.1}{49}} = 8.44$

We are given the following in the question:

Population mean, μ = 106

Sample mean, $\bar{x}$ = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 106\\H_A: \mu > 106$

We use one-tailed(right) t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135$

Now,

$t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833$

Since,

$t_{stat} > t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

4 0

3 years ago

Someone help me asap pls​

Someone help me asap pls