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2 years ago

Which statement is true

Mathematics

2 answers:

deff fn [24]2 years ago

6 0

I would say the answer is D.

Jobisdone [24]2 years ago

4 0

The answer is point D.)
m∠2=m∠4

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I need help please.

SIZIF [17.4K]

Answer:

A because there isnt a mode, hope this helps\

3 0

2 years ago

Solve the inequality Z-2>2 Z > 2?

Troyanec [42]

Answer:

Z> 4

So Z> 2 is true

Step-by-step explanation:

Given that:

Z-2>2

Adding 2 on both sides

Z-2 + 2 >2 + 2

Z > 4

hence the equation Z > 2 is true

i hope it will help you!

5 0

2 years ago

Subtract. –84 – 55 a. –29 b. –139 c. 29 d. 139

Hitman42 [59]

-84 - 55 = -139 Here both have same sign (-) so u add and put the common sign(-) .... So the answer is option B ( -139)... Hope it helps!!!

8 0

2 years ago

Circle A has been enlarged to create circle A'. The table below shows the circumference of both circles.

stich3 [128]

Answer:

Option (2) 1.5

Step-by-step explanation:

Rule for the dilation of an image by a scale factor is,

Scale factor = $\frac{\text{Radius of image circle A'}}{\text{Radius of pre-image A}}=\frac{r'}{r}$

= $\frac{(2\pi) r'}{(2\pi) r}$

= $\frac{\text{Circumference of image}}{\text{Circumference of pre-image}}$

= $\frac{10.5}{7}$

= 1.5

Therefore, scale factor was used to create a circle A' = 1.5

Option (2) will be the correct option.

8 0

2 years ago

4. In a sample of 1000 people, 130 can wiggle their ears. Two unrelated people are

Elina [12.6K]

Using the hypergeometric distribution, it is found that there is a 0.7568 = 75.68% probability that neither can wiggle his or her ears.

The people are chosen from the sample without replacement, which is why the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.

In this problem:

1000 people means that $N = 1000$
130 can wiggle their ears, thus $k = 130$
Two are selected, thus $n = 2$ .

The probability that neither can wiggle his or her ears is P(X = 0), thus:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = x) = h(0,1000,2,130) = \frac{C_{130,0}C_{870,2}}{C_{1000,2}} = 0.7568$

0.7568 = 75.68% probability that neither can wiggle his or her ears.

A similar problem is given at brainly.com/question/24826394

5 0

1 year ago