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3 years ago

Please hellllp!!! pleaseeeeee

Mathematics

1 answer:

kkurt [141]3 years ago

3 0

The answer is D because if the perimeter of wxyz can be found by multiplying 2 with the perimeter of hjkl, then the area would be area of hjkl times 2

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The population of a certain geographic region is approximately 103 million and grows continuously at a relative growth rate of 1

aev [14]

The population increases by a factor 1.0167 every year so the require figure is:
103 * 1 . 0167^5 = 112 million

3 0

3 years ago

Sylvie finds the solution to the system of equations by graphing. y = x + 1 and y = x – 1 Which graph shows the solution to Sylv

sesenic [268]

Hey user☺☺

The graph will have no solution.

Hope it works

6 0

3 years ago

Read 2 more answers

Help meeeeeeeeeeeeeeeweeeeeeeeeeeeeeeee

Yuri [45]

Step-by-step explanation:

This can be put into fractions like this

Say x is the length of the missing dimension

8.5/11 = 10.5/x

Because the 2 drawings are SIMILAR, the ratio of their side lengths has to also be the same

8.5 ÷ 10.5 = 0.8095

11 × 0.8095 = 8.9047

Roung the answer to the nearest tenth

x = 8.9 in

5 0

3 years ago

-4. (5 + 1) -3 + 3+ 2

antiseptic1488 [7]

Answer:

-13

Step-by-step explanation:

7 0

2 years ago

For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12

notsponge [240]

The binomial distribution is given by,
P(X=x) = $(^{n}C_{x})p^{x} q^{n-x}$
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability <span>of obtaining a score less than or equal to 12.
</span>∴ P(X≤12) = $(^{100}C_{x})(0.2)^{x} (0.8)^{100-x}$
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) = $(^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1}$ + $(^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3}$ + $(^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5}$ + $(^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7}$ + $(^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9}$ + $(^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11}$ + $(^{100}C_{12})(0.2)^{12} (0.8)^{100-12}$

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.

7 0

3 years ago