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3 years ago

Find the endpoint of the line segment with the given endpoint and midpoint. Endpoint:(2,1) midpoint:(0,4)

Mathematics

1 answer:

Alex Ar [27]3 years ago

4 0

Answer: (-2, 7)

Step-by-step explanation:

X started at 2 and went to the left 2 to get to the mid point of 0. Go 2 more to the left to get to the other end of -2.

Y started at 1 and went up 3 to get to the midpoint of 4. Go up another 3 to get to the endpoint of 7.

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Legit question what is 10x 10 100 or 1000 I keep getting confused isn't it 100?

s344n2d4d5 [400]

Yea it is 100 :)

10x10=100

7 0

3 years ago

Read 2 more answers

Whats is the answer to 5/8 -1/10

LuckyWell [14K]

Answer:

Hello, Your answer will be 45

Step-by-step explanation:

5+8=5*+%=45 <em>Hope This Helps!</em>

5 0

3 years ago

Find all the missing elements : C=120degrees<br> b= 5<br> c=11

mezya [45]

Answer:

A = 36.8°

B = 23.2°

a = 7.6

Step-by-step explanation:

Given:

C = 120°

b = 5

c = 11

Required:

Find A, B, and a.

Solution:

✔️To find B, apply the Law of Sines

$\frac{sin(B)}{b} = \frac{sin(C)}{c}$

Plug in the values

$\frac{sin(B)}{5} = \frac{sin(120)}{11}$

Cross multiply

Sin(B)*11 = sin(120)*5

Divide both sides by 11

$sin(B) = \frac{sin(120)*5}{11}$

Sin(B) = 0.3936

B = $sin^{-1}(0.3936)$

B = 23.1786882° ≈ 23.2° (nearest tenth)

✔️Find A:

A = 180° - (B + C) (sum of triangle)

A = 180° - (23.2° + 120°)

A = 36.8°

✔️To find a, apply the Law of sines:

$\frac{sin(A)}{a} = \frac{sin(B)}{b}$

Plug in the values

$\frac{sin(36.8)}{a} = \frac{sin(23.2)}{5}$

Cross multiply

a*sin(23.2) = 5*sin(36.8)

Divide both sides by sin(23.2)

$a = \frac{5*sin(36.8)}{sin(23.2)$

a = 7.60294329 ≈ 7.6 (nearest tenth)

5 0

3 years ago

Suppose that in an alternate universe, the gambler's fallacy is true: the more a gambler loses, the more likely she is to win th

stepladder [879]

Answer and explanation:

The gambler's fallacy is the fallacy of belief that if an event such as a loss occurs more frequently in the past, it is less likely to happen in the future. We assume here that this belief is true, therefore

If she loses, her probability of winning increases =3/4

If she wins, her probability to win is normal =1/2

Given that probability of winning is 1/2

Probability of losing is 1-1/2=1/2

Probability that she wins the tournament is probability that she wins the first two games and loses the last or wins the first game, loses the second and wins the last or loses the first game and wins the last two games or probability that she wins all three games

=1/2*1/2*1/2+1/2*1/2*3/4+1/2*3/4*1/2+1/2*1/2*1/2

=25/48

Probability of winning the tournament if she loses the first game

=1/2*3/4*1/2= 3/16

Note: whenever there is "or" in probability, you add

4 0

3 years ago

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

4 0

3 years ago