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2 years ago

7

Given △ABC, use a dilation with the center at the origin to make a similar triangle with side lengths three times as large. What

are the coordinates of C' of the image
(–6, 3)

(–3, 6)

(3, –6)

(6, –3)

2 answers:

Artist 52 [7]2 years ago

7 0

The image of the dilation is shown below, with the centre of dilation (0,0) and scale factor of 3

The coordinate of C' is (6, -3) which is three times of the coordinate of C(2, -1)

Afina-wow [57]2 years ago

6 0

Answer:

6,-3

Step-by-step explanation:

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The average life of a bread-making machine is 7 years, with a standard deviation of 1 year. Assuming that the lives of these mac

Alina [70]

Answer:

a) $P(6.4$

b) $a=7 +1.036*0.333=7.345$

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable life of a bread making machine. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =7,\sigma =1)$

We take a sample of n=9 . That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=7, \frac{1}{\sqrt{9}})$

Solution to the problem

Part a

(a) the probability that the mean life of a random sample of 9 such machines falls between 6.4 and 7.2

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

The standard error is given by this formula:

$Se=\frac{\sigma}{\sqrt{n}}=\frac{1}{\sqrt{9}}=0.333$

We want this probability:

$P(6.4$

Part b

b) The value of x to the right of which 15% of the means computed from random samples of size 9 would fall.

For this part we want to find a value a, such that we satisfy this condition:

$P(\bar X>a)=0.15$ (a)

$P(\bar X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.036. On this case P(Z<1.036)=0.85 and P(Z>1.036)=0.15

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.036$

And if we solve for a we got

$a=7 +1.036*0.333=7.345$

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

8 0

2 years ago

A population has a standard deviation of 80. A random sample of 400 items from this population is selected. The sample mean is d

Helga [31]

Answer:

$ME= 1.8808 * \frac{80}{\sqrt{400}} =7.5232$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma =80$ represent the population standard deviation

n=400 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Since the Confidence is 0.94 or 94%, the value of $\alpha=0.06$ and $\alpha/2 =0.03$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that $z_{\alpha/2}=1.8808$

The margin of error is given by:

$ME= 1.8808 * \frac{80}{\sqrt{400}} =7.5232$

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2 years ago

Which of the following are exterior angles? Check all that apply.

Helen [10]

An exterior angle of a triangle is an angle formed by one side of the triangle and the extension of an adjacent side of the triangle.

FACTS:

Every triangle has 6 exterior angles, two at each vertex.
Notice that the "outside" angles that are "vertical" to the angles inside the triangle are NOT called exterior angles of a triangle.

1. Angles 1, 5 and 6 are inteior angles of the given triangle.

2. Angle 3 is vertical to the interior angle 5.

3. Angles 2 and 4 are exterior angles of the triangle.

Answer: correct options are A and B.

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2 years ago

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A 6-kg bag of fertilizer covers a rectangular area of 2.5 m by 5.0 m.

Naily [24]

Answer

Given,

Mass of bag = 6 Kg

Area covered by the bag = 2.5 x 5 = 12.5 m²

a) Bags required to cover area = 7.5 x 10.2

= 76.5 m²

Number of bags, N = $\dfrac{76.5}{12.5}$

N = 6.12 = 7 (approx.)

7 Bags of 6 Kg fertilizers will be needed to cover the area.

b) Cost of 6 Kg fertilizers = $15.50

Cost of 7 bags of 6 Kg fertilizers = 7 x $15.50

= $108.50.

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2 years ago

The table shows different ways that Cameron can display his 12 model cars on shelves. How many shelves will display 2 cars if 8

EleoNora [17]

2 shelves will display 2 cars because if 8 shelves display 1 car then there would be missing 4 cars to display and if you display 2 cars in one shelf you will need 2 shelves

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2 years ago

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