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larisa [96]
3 years ago
15

Select the correct answer from each drop-down menu. In the figure, x = , y = , and z = .

Mathematics
2 answers:
alexdok [17]3 years ago
8 0
X= 95
z=180
y=147

based on from making sure each triangle added up to 180 degrees and with the outer angles
Gekata [30.6K]3 years ago
7 0

Answer:

x=95

y=147

z=128

Step-by-step explanation:

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What is the range of f(x)=1/2^x
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Downloaded Photomath it’s going to help you a lot
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PLEASE HELP ME i DONT KNOW how to do this!​
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It's b

Step-by-step explanation:

It is b or the second option because -2 is equal or greater than A therefore it is option 2

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3 years ago
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a farmer uses 1/3 of his land to plant cassava 3/5 of the remaining land to plant Maize and the rest vegetables if vegetables co
Rzqust [24]

Answer:

37.5 acres.

Step-by-step explanation:

Fraction of his land  where vegetables are grown

= 1 - 1/3 - 2/3 *3/5

(after taking away 1/3 we get 2/3 and the farmer uses 3/5 of this 2/3 to grow Maize.

= 1 - 1/3 - 6/15              LCD of 3 and 15= 15  so we write:

= 1 - 5/15 - 6/15

= 1 - (5+6)/16

=  1 - 11/15

= 4/15.

4 /15 is equivalent to  10 acres

so  total  area (= 1)  =   10 / 4/15     Invert the 4/15 and multiply:

= 10 * 15/4

=  150/4

= 37.5 acres.

3 0
3 years ago
What is the difference between 12/8 and 3/4?
Katen [24]

Answer:

3/4

Step-by-step explanation:

Difference means subtraction

12/8 - 3/4

We need a common denominator of 8

12/8 - 3/4 * 2/2

12/8  - 6/8

 6/8

Divide by 2 in the numerator and denominator

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6 0
3 years ago
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A car insurance company has determined that 9% of all drivers were involved in a car accident last year. among 14 drivers living
andriy [413]

The probability of getting 3 or more who were involved in a car accident last year is 0.126.

Given 9% of the drivers were involved in a car accident last year.

We have to find the probability of getting 3 or more who were involved in a car accident last year if 14 were selected randomly.

We have to use binomial theorem which is as under:

nC_{r}p^{r} (1-p)^{n-r}

where p is the probability an r is the number of trials.

Probability that 3 or more involved in a car accident last year if 14 are randomly selected=1-[P(X=0)+P(X=1)+P(X=2)]

=1-{14C_{0}0.9^{0} (0.1)^{14} +14C_{1} (0.9)^{1} (0.1)^{13} +14C_{2} (0.9)^{2} (0.1)^{12}}

=1-{1*0.2670+14!/13!*0.9*0.29+14!/2!12!*0.0081*0.2358}

=1-{0.2670+0.3654+0.2358}

=1-0.8682

=0.1318

Among the options given the nearest is 0.126.

Hence the probability that 3 or more are involved in the accident is 0.126.

Learn more about probability at brainly.com/question/24756209

#SPJ4

7 0
2 years ago
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