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2 years ago

A triangle with vertices (0,0), (3,6), (7,4) is dilated by a factor of 3 with its center at the origin.

Mathematics

1 answer:

Mariana [72]2 years ago

7 0

Answer:

$A' = (0,0)$

$B'= (9,18)$

$C'= (21,12)$

Step-by-step explanation:

Given

$A= (0,0)$

$B= (3,6)$

$C= (7,4)$

$k = 3$ -- dilation factor

Required

Determine the coordinates of the new triangle

This is calculated using:

$New = Old * k$

So, we have:

$A' = A *3$

$A' = (0,0) *3$

$A' = (0,0)$

$B' = B * 3$

$B'= (3,6)*3$

$B'= (3*3,6*3)$

$B'= (9,18)$

and

$C' = C * 3$

$C'= (7,4)*3$

$C'= (7*3,4*3)$

$C'= (21,12)$

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One urn contains one blue ball (labeled b1) and three red balls (labeled r1, r2, and r3). a second urn contains two red balls (r

Flura [38]

Answer:

12 possibilities

Step-by-step explanation:

In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.

The same thing occurs in the second urn, as all balls have different labels.

The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).

For the first urn, we have a combination of 4 choose 2:

C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities

For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.

In total we have 6 + 6 = 12 possibilities.

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2 years ago

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A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,

Andru [333]

Answer:

A. Null and alternative hypothesis:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848$

The degrees of freedom for this sample size are:

$df=n-1=42-1=41$

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>4.848)=0.00002$

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.02 \cdot 173.283=350$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190$

The 95% confidence interval for the mean difference is (490, 1190).

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2 years ago

A locker requires a three-digit code to open the lock. The code must contain one letter and two numbers, and no letter or number

mixas84 [53]

A locker requires a three-digit code to open the lock. The code must contain one letter and two numbers, and no letter or number can be repeated. You can choose from among four letters, A, B, C, and D, and two numbers, 5 and 6.

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If a code is chosen at random, the probability that it has a letter that immediately follows an odd number is: (1/8, 1/6, 1/3, 2/5, 2/3)

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1 year ago

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lidiya [134]

Answer:

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2 years ago

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soldi70 [24.7K]

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1 year ago