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3 years ago

triangle fdp is reduced with a scale factor of 1/2 and a center of (0,0). Find the coordinates of the new vertices.

Mathematics

1 answer:

Mkey [24]3 years ago

4 0

The new points are:
P' (-3, 0)
F' (1, 2)
D' (0, -2.5)

The shape is being transformed with a dilation of a scale factor of 1/2. This means that we take the given points and multiply them by 1/2 to find the new points.

Look on the graph for the current points and multiply each one by 1/2. You will have the new points.

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A student has some $1 and $5 bills in his wallet. He has a total of 14 bills that are worth $34. How many of each type of bil

jeka57 [31]

Solution:

we are given that

A student has some $1 and $5 bills in his wallet.

Let there x $1 bills and y number of $5 bills.

He has a total of 14 bills that are worth $34.

So we can write

$x+y=14$

$x+5y=34$

Now solve these two equations together using substitution as follows

$14-y+5y=34\\
\\
4y=34-14\\
\\
4y=20\\
\\
y=5\\
\\
x=14-y=14-5=9\\$

Hence there are 9 $1 bills and 5 bills are of $5.

5 0

3 years ago

Consider the computer output below. Fill in the missing information. Round your answers to two decimal places (e.g. 98.76). Test

slamgirl [31]

Answer:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

$t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124$

The 95% confidence interval would be given by (96.625;100.915)

a) $df=n-1= 19-1= 18$

b) $p_v =2*P(t_{18}$

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

c) The only thing that changes is the p value and would be given by:

$p_v =P(t_{18}>-1.124)=0.862$

But again since the p value is higher than the significance level we fail to reject the null hypothesis.

Step-by-step explanation:

Previous concepts and data given

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".

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".

$\bar X=98.77$ represent the sample mean

$s=4.77$ represent the sample standard deviation

n=19 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if we have significant difference on the mean of 100, the system of hypothesis would be:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

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

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

And we can calculate the Standard error for the mean like this:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

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 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$98.77-1.96\frac{4.77}{\sqrt{19}}=96.625$

$98.77+1.96\frac{4.77}{\sqrt{19}}=100.915$

So on this case the 95% confidence interval would be given by (96.625;100.915)

Part a

The degree of freedom are given by:

$df=n-1= 19-1= 18$

Part b

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124$

Then since is a two sided test the p value would be:

$p_v =2*P(t_{18}$

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Part c

If the system of hypothesis on this case are:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu > 100$

The only thing that changes is the p value and would be given by:

$p_v =P(t_{18}>-1.124)=0.862$

But again since the p value is higher than the significance level we fail to reject the null hypothesis.

7 0

3 years ago

A ball is thrown into the air with an initial velocity of 22 meters per second. The quadratic function h(x)= -4.9t square +22t+5

MrRa [10]

Answer:

Height of the ball after 3 seconds = 27.4 feet.

Step-by-step explanation:

h(t) = -4.9t^2 + 22t + 5.5

h(3) means the height of the ball at time 3 seconds after the ball was thrown.

h(3) = -4.9(3)^2 + 22(3) + 5.5

= 27.4 feet

5 0

3 years ago

What is the equation of the following line written in general form?

AlekseyPX

Answer:

add 1;6 and 6,2 hope this helps ;)

8 0

3 years ago

If the length of the radius of a sphere is multiplied by 4, what is the effect on the volume?

ryzh [129]

The volume increases.

5 0

3 years ago