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3 years ago

Find each measurement indicated. Round your answer as the nearest tenth. HELP PLEASE!!!!!! URGENT

Mathematics

1 answer:

MaRussiya [10]3 years ago

8 0

Answer:

(1) length AB = 21.0

(2) length BC = 45.0

Step-by-step explanation:

(1) To determine AB, apply Cosine rule;

$|AB|^2 = |AC|^2 + |BC|^2 \ - \ 2[|AC| \times |BC]cos \ C\\\\|AB|^2 = 13^2 + 29^2 \ - \ 2(13 \times 29) cos(41)\\\\|AB|^2 = 1010 - 569.05\\\\|AB|^2 = 440.95\\\\|AB|= \sqrt{ 440.95} \\\\|AB| = 21.0$

(2) To determine BC, also apply cosine rule;

$|BC|^2 = |AB|^2 + |AC|^2 \ - \ 2[|AB| \times |AC]cos \ A\\\\|BC|^2 = 30^2 + 21^2 \ - \ 2(30 \times 21) cos(123)\\\\|BC|^2 = 1341 - (-686.245)\\\\|BC|^2 = 1341 + 686.245\\\\ |BC| = 2027.245\\\\|BC|= \sqrt{ 2027.245} \\\\|BC| = 45.0$

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Need help on this if anyone can

Lunna [17]

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3 years ago

I really need help with this section

Tpy6a [65]

Looks like you did most of it.

The next step would be setting them equal to 105, like the problem suggests.

$5z+3x+7x=105$

$15x=105$

$x=7$

Then we can plug back in to find the numbers.

$5*7=35$

$3*7=21$

$7*7=49$

$35+21+49=105$

The three numbers are $35, 21, 49$ .

Hope this helps.

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4 years ago

Read 2 more answers

Jane must get at least three of the four problems on the exam correct to get an A. She has been able to do 80% of the problems o

NISA [10]

Answer:

a) There is n 81.92% probability that she gets an A.

b) If she gets the first problem correct, there is an 89.6% probability that she gets an A.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the answer is correct, or it is not. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

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

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

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

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

And $\pi$ is the probability of X happening.

For this problem, we have that:

The probability she gets any problem correct is 0.8, so $\pi = 0.8$ .

(a) What is the probability she gets an A?

There are four problems, so $n = 4$

Jane must get at least three of the four problems on the exam correct to get an A.

So, we need to find $P(X \geq 3)$

$P(X \geq 3) = P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 3) = C_{4,3}.(0.80)^{3}.(0.2)^{1} = 0.4096$

$P(X = 4) = C_{4,4}.(0.80)^{4}.(0.2)^{0} = 0.4096$

$P(X \geq 3) = P(X = 3) + P(X = 4) = 2*0.4096 = 0.8192$

There is n 81.92% probability that she gets an A.

(b) If she gets the first problem correct, what is the probability she gets an A?

Now, there are only 3 problems left, so $n = 3$

To get an A, she must get at least 2 of them right, since one(the first one) she has already got it correct.

So, we need to find $P(X \geq 2)$

$P(X \geq 3) = P(X = 2) + P(X = 3)$

$P(X = 2) = C_{3,2}.(0.80)^{2}.(0.2)^{1} = 0.384$

$P(X = 4) = C_{3,3}.(0.80)^{3}.(0.2)^{0} = 0.512$

$P(X \geq 3) = P(X = 2) + P(X = 3) = 0.384 + 0.512 = 0.896$

If she gets the first problem correct, there is an 89.6% probability that she gets an A.

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4 years ago

Hannah is 16 years younger than Ericka. In 5 years Ericka will be twice as old as Hannah. How old is each

vazorg [7]

Answer:11 and 27

Step-by-step explanation:

11+16=27 and they are 16 years apart and because it says 5 years 27+5=32 and if you add 5 to 11 you get 16 and 16+16=32

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2 years ago

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Explain how to write and evaluate an algebraic expression that represents the situation. The blueberries were divided equally in

ser-zykov [4K]

The variable is the number of blueberries, the operation is division, and the constant is 6. Write the division expression as the fraction, b/6 . To evaluate, substitute 210 in for the variable, b, and simplify. 210 divided by 6 is 35. So each bowl has 35 blueberries.

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3 years ago

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