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Svet_ta [14]
2 years ago
10

4 12 36 108 n ? whats after n

Mathematics
1 answer:
lilavasa [31]2 years ago
5 0

Answer:

324

Step-by-step explanation:

4x3=12

12x3=36

36x3=108

108x=324

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Whats the answer to this please
anastassius [24]
Y = 3x
so
if x = 5 then y = 15
if x = 10 then y = 30

so a = 10 and b = 15

answer is B. second choice
a = 10 and b = 15
3 0
3 years ago
Can anyone help me with this math problem please?
aleksklad [387]

Answer:

I said, certified freak

Seven days a week

Wet-as* pus

Step-by-step explanation:

6 0
2 years ago
If R is the midpoint of QS, RS= 2x - 4, ST = 4x-1, and RT = 8x - 43, find QS.
Pepsi [2]

Answer: QS=68 units

Step-by-step explanation:

<em>RT=RS+ST</em>

<em>8x-43=2x-4+4x-1</em>

<em>8x-43=6x-5</em>

<em>8x-43-6x=6x-5-6x</em>

<em>2x-43=-5</em>

<em>2x-43+43=-5+43</em>

<em>2x=38</em>

<em>Divide both parts of the equation by 2:</em>

<em>x=19</em>

<em>QS=RS+RQ</em>

<em>As, RS=RQ</em>

<em>QS=2RS</em>

<em>QS=(2)((2)(19)-4))</em>

<em>QS=(2)(38-4)</em>

<em>QS=(2)(34)</em>

<em>QS=68</em>

6 0
1 year ago
Assume that when adults with smartphones are randomly​ selected, 58​% use them in meetings or classes. If 10 adult smartphone us
Tasya [4]

Answer:

79.85% probability that at least 5 of them use their smartphones in meetings or classes.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they use their smartphone during meetings or classes, or they do not. The probability of an adult using their smartphone in these situations are independent of other adults. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Assume that when adults with smartphones are randomly​ selected, 58​% use them in meetings or classes.

This means that p = 0.58

10 adults selected.

This means that n = 10

Find the probability that at least 5 of them use their smartphones in meetings or classes.

P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{10,5}.(0.58)^{5}.(0.42)^{5} = 0.2162

P(X = 6) = C_{10,6}.(0.58)^{6}.(0.42)^{4} = 0.2488

P(X = 7) = C_{10,7}.(0.58)^{7}.(0.42)^{3} = 0.1963

P(X = 8) = C_{10,8}.(0.58)^{8}.(0.42)^{2} = 0.1017

P(X = 9) = C_{10,9}.(0.58)^{9}.(0.42)^{1} = 0.0312

P(X = 10) = C_{10,10}.(0.58)^{10}.(0.42)^{0} = 0.0043

P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.2162 + 0.2488 + 0.1963 + 0.1017 + 0.0312 + 0.0043 = 0.7985

79.85% probability that at least 5 of them use their smartphones in meetings or classes.

3 0
3 years ago
On two examinations, you have grades of 87 and 86. There is an optional final examination, which counts as one grade. You decide
noname [10]

Step-by-step explanation:

We have

87+86+x

T is the total scores

n is number of tests done = 3

T/n is greater than or equal to 90

To get minimum test

T/n = 90

T = 90*n

= 90n

We solve for x

Remember n = 3

90x3 = 270

270 = 87+86+x

270 = 173+x

X = 270-173

X = 97

1.

X = 97

2.

You didn't add a value in your part b question. I used 80 though.

T/n < 80

T<80n

80x3= 240

T<240

87+86+x<240

X<240-173

X<67

X has to be less than 67 so we conclude that if x <=67 b grade will be lost.

3 0
3 years ago
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