What is the value of a? 06√2 9 O 8√3 O 36√2

Mathematics

1 answer:

melamori03 [73]2 years ago

7 0

Answer: 6 $\sqrt{2}$

Step-by-step explanation: why the process of elimination is because 9 is sticking like a sore thumb and the last one is too big the first one and the third one are similar but the third one has a square root of 3 so it would be THE FIRST ONE.

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use the formula for the area of a trapezoid A=h(b1+b2/2), were area is A b1 and b2 are the length of the base and h is the hight

Colt1911 [192]

Answer: 300 yards

Step-by-step explanation:

Since b1 is equal to 14 yds and b2 is equal to 26 yds we can plug that into the equation below (in the image).

b1 + b2 = 14 + 26 = 40 yards

Now that we have found (b1 + b2), you can multiply that by h (the height of the trapezoid) which is 15, so we get 40 x 15 = 600.

In a lot of area equations, you will have to divide by two like in the trapezoid equation. Divide our number (600) by 2, so we get 600 ÷ 2 = 300.

300 yds is your final answer.

Hope this helps!

5 0

4 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$