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kari74 [83]
4 years ago
14

Which is the pair of congruent

Mathematics
1 answer:
Maksim231197 [3]4 years ago
5 0

Answer:

CBA = DEA

Step-by-step explanation:

Theyre the only right angles listed.

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Permutations vs combinations(one step):
Bogdan [553]

Answer:

the expression is: ^{15}P_3=\frac{15!}{(15-3)!}

and 2730 different ways are possible.

Step-by-step explanation:

In the given question we have to choose a president, vice

president, and secretary.

Permutations are used when order is important so, as in the given question order is required. We would use permutations

The formula used is: ^nP_r=\frac{n!}{(n-r)!} where:

n=total number of objects

r=number of objects selected

In our case: n=15, r=3

Putting values and finding the answer:

^nP_r\\^{15}P_3=\frac{15!}{(15-3)!} =2730 \ ways

So, the expression is: ^{15}P_3=\frac{15!}{(15-3)!}

and 2730 different ways are possible.

7 0
3 years ago
-7 = -c - 29 Easy Points
Luden [163]

Hi there!  

»»————- ★ ————-««

I believe your answer is:  

c = - 22

»»————- ★ ————-««  

Here’s why:  

⸻⸻⸻⸻

\boxed{\text{Solving for 'c'...}}\\\\-7 = -c-29\\-------------\\\rightarrow -c -29 = -7\\\\\rightarrow -c - 29 + 29 = -7 + 29\\\\\rightarrow -c  = 22\\\\\rightarrow\frac{-c=22}{-1}\\\\\rightarrow \boxed{c = -22}

⸻⸻⸻⸻

»»————- ★ ————-««  

Hope this helps you. I apologize if it’s incorrect.  

5 0
3 years ago
A scientist claims that 4% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of
oksano4ka [1.4K]

Answer:

The probability that the proportion of airborne viruses in a sample of 662 viruses would be greater than 6%=0.00427

Step-by-step explanation:

We are given that

\mu_{\hat{p}}=p=4%=0.04

n=662

We have to find the probability that the proportion of airborne viruses in a sample of 662 viruses would be greater than 6%.

q=1-p=1-0.04=0.96

\sigma_{\hat{p}}=\sqrt{p(1-p)/n}

\sigma_{\hat{p}}=\sqrt{\frac{0.04(1-0.04)}{662}}

\sigma_{\hat{p}}=0.0076

Now,

P(\hat{p}>0.06)=1-P(\hat{p}

=1-P(\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}

=1-P(Z

=1-0.99573

P(\hat{p}>0.06)=0.00427

Hence, the probability that the proportion of airborne viruses in a sample of 662 viruses would be greater than 6%=0.00427

7 0
3 years ago
Someone solve for me please WILL MARK THE BRAINLIEST FOR EXPLANATION !!!!!
Harrizon [31]
The answer to the question

7 0
4 years ago
Special Right Triangle
MA_775_DIABLO [31]
<h3>let ABC be a triangle</h3>

where,

  • AB= 6cm
  • BC= a
  • AC = b

now

applying tan 30°

tan theta = perpendicular/base

=>tan 30° = 6/a

(since tan 30° = 1/√3)

=>1/√3 = 6/a

=> a = 6√3

applying cos Q for finding b

=>cos 60° = b/h

(since cos 60° = 1/2)

=>1/2= 6/b

=> b= 12

3 0
3 years ago
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