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2 years ago

4

Consider reflections of δjkl. what line of reflection maps point k to point k' at (–5, 2)? what line of reflection maps point l

to point l' at (–2, 3)?

2 answers:

mr_godi [17]2 years ago

8 0

Answer:

1st one is y-axis. 2nd one is y=-x.

Step-by-step explanation:

I got it right on Edge.

myrzilka [38]2 years ago

5 0

Its y-axis for the first answer and the other one its, y=-x

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Bob is having a party and want to order some cake. there will be 57 guest and he wants to make sure that everyone gets a slice o

Dafna1 [17]

To find out which cake is better for the price, we need to multiply the cake amounts by how many cakes of that kind we need. You need 5 20cm cakes, so we multiply 5 by the price of the 20cm cake (13.50) to get $67.50. You need 3 25cm cakes, so 3 x 18.75 = 56.25.

20cm cake: $67.50
25cm cake: $56.25

Answer: The 25cm cake is cheaper for more people.

4 0

2 years ago

Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The

Harlamova29_29 [7]

Answer:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

We have the following data given:

$n_1 =20$ represent the sample size for group 1

$n_2 =20$ represent the sample size for group 2

$\bar X_1 =43.5$ represent the sample mean for the group 1

$\bar X_2 =40.1$ represent the sample mean for the group 2

$s_1=4.1$ represent the sample standard deviation for group 1

$s_2=3.2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525$

And the deviation would be just the square root of the variance:

$S_p=3.678$

The statistic is givne by:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0

2 years ago

Name the postulate or theorem you can use to prove ∆ABD ≅ ∆CBD.

lozanna [386]

Hey there!

Triangle CBD is congruent to triangle ABD. This means that BD is congruent to BD by reflexive property. You have two congruent angles, and one congruent side. This would be AAS theorem. The answer is D.

I hope this helps!

5 0

2 years ago

Read 2 more answers

The nth term of a sequence is 20 - n2

maks197457 [2]

Answer:

11

Fifth term

Step-by-step explanation:

The third term is:

20 − (3)² = 11

If the term is negative:

20 − n² < 0

n² > 20

n > √20

n > 4.47

The first term to have a negative value is the fifth term.

5 0

2 years ago

Rework problem 9 from section 3.2 of your text, involving independent and disjoint events. For this problem, assume that Pr[A∪B]

tatuchka [14]

Two events are said to be Disjoint or Mutually Exclusive if the two events can not happen at the same time.For example when we throw a die getting an even number is disjoint to getting an odd number.

I.e Probability(A∩B)=0

Let me explain this concept through venn diagram.

Pr[A∪B]=0.7, Pr[A]=0.25

Since events are Disjoint

Pr[A∩B]=0

Pr[A∪B]=Pr[A] + Pr[B]

0.7=0.25 +Pr[B]

0.7-0.25=Pr[B]

⇒Pr[B]=0.45=45/100=9/20

Now events are said to be independent if Pr[A and B]=Pr[A] ×Pr[B]

Events are said to be independent if occurrence of one is not affected by occurrence of other.For example getting multiple of 2 as one event and getting multiple of 3 as second event when we throw a die.

Pr[A∪B]=0.7, Pr[A]=0.25

Pr[A∪B]= Pr[A]+ Pr[B]-Pr[A∩B]

But Pr[A∩B]= Pr[A] ×Pr[B]

⇒Pr[A∪B]= Pr[A]+ Pr[B]- Pr[A] ×Pr[B]

⇒0.7=0.25+p-0.25×p

⇒0.7-0.25=p- 0.25 p

⇒0.45=0.75 p

⇒p= 0.45/0.75

⇒p =3/5

5 0

2 years ago