What is the image of F for a dilation with center (0, 0) and a scale factor of 10?

Which glide reflection describes the mapping ABC DEF

almond37 [142]

Answer:

1. Reflect ABC about the line AC and then translate 1 unit to the right.

2. Translate ABC 1 unit to the right and then reflect it about the line AC.

Step-by-step explanation:

We are given that,

ABC is transformed using glide reflection to map onto DEF.

Since, we know,

'Glide Reflection' is the transformation involving translation and reflection.

So, we can see that,

ABC can be mapped onto DEF by any of the following glide reflections:

1. Reflect ABC about the line AC and then translate 1 unit to the right.

2. Translate ABC 1 unit to the right and then reflect it about the line AC.

Hence, any of the two glide reflection will map ABC onto DEF.

6 0

3 years ago

Read 2 more answers

The CEO of the Jen and Benny's ice cream company is concerned about the net weight of ice cream in their 50 ounce ice cream tubs

MissTica

Answer:

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

We need to find the degrees of freedom given by:

$df = n-1= 26-1 = 25$

Since is a right tailed test the p value would be:

$p_v =P(t_{25}>1.8025)=0.0418$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=54.98$ represent the sample mean

$s=8.43$ represent the sample deviation

$n=26$ sample size

$\mu_o =52$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 52, the system of hypothesis would be:

Null hypothesis: $\mu \leq 52$

Alternative hypothesis: $\mu > 52$

Since we don't know the population deviation, 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".

Calculate the statistic

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

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

P-value

We need to find the degrees of freedom given by:

$df = n-1= 26-1 = 25$

Since is a right tailed test the p value would be:

$p_v =P(t_{25}>1.8025)=0.0418$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.

7 0

3 years ago

Read 2 more answers

Show that (x-5) is a factor of x^3-3x^2-13x+15

mario62 [17]

Answer:

sorry not in that grade

Step-by-step explanation:

6 0

3 years ago

Find the other endpoint of the line segment with the given endpoint and midpoint. Endpoint: (-2, -6) Midpoint: (-5, 1)

FromTheMoon [43]

Answer:

(-8,8)

Step-by-step explanation:

The midpoint $M(x_M,y_M)$ of the segment AB with endpoints $A(x_A,y_A)$ and $B(x_B,y_B)$ has coordinates

$x_M=\dfrac{x_A+x_B}{2}\\ \\y_M=\dfrac{y_A+y_B}{2}$

In your case, $A(-2,-6), \ M(-5,1),$ then

$-5=\dfrac{-2+x_B}{2}\Rightarrow -2+x_B=-10,\ x_B=-10+2=-8\\ \\1=\dfrac{-6+y_B}{2}\Rightarrow -6+y_B=2,\ y_B=2+6=8$

3 0

3 years ago

#12 with explanation please

4vir4ik [10]

Answer:

30.01 units

Step-by-step explanation:

Given vertices WXYZ:

W(-4,5), X(2,4), Y(3, -4), Z(-1, -6)

Perimeter is the sum of the side length.

Use distance formula to find each side:

WX = $\sqrt{(2+4)^2+(4-5)^2} =\sqrt{37}$ ≈ 6.08
XY = $\sqrt{(3-2)^2+(-4-4)^2} =\sqrt{65}$ ≈ 8.06
YZ = $\sqrt{(-1-3)^2+(-6+4)^2} =\sqrt{20}$ ≈ 4.47
WZ = $\sqrt{(-1+4)^2+(-6-5)^2} =\sqrt{130}$ $\sqrt{(-1+4)^2+(-6-5)^2}=\sqrt{130}$ ≈ 11.40

Find perimeter:

P = 6.08 + 8.06 + 4.47 + 11.40 = 30.01 units

7 0

3 years ago