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2 years ago

Find the value of x. Round your answer to the nearest tenth.

Mathematics

1 answer:

NARA [144]2 years ago

4 0

Answer:

24.2

Step-by-step explanation:

c = b·sin(C)/sin(B) = 24.15071

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Someone plzzz help I’m lost :(

JulsSmile [24]

Answer:

j = (h-4) / k

Step-by-step explanation:

Unless there is more information given about this question somewhere else in the assignment, I'm assuming you are being asked to solve this equation for j.

If so, then its just a matter of rearranging the equation so that it is in the form

j = (expression containing other variables)

(h-4) / j = k (multiply both sides by j)

(h-4) = jk (divide both sides by k)

j = (h-4) / k

3 0

3 years ago

Simplify the expression

AlexFokin [52]

Answer:

Step-by-step explanation:

$\frac{3+4\iota}{2+4\iota}=\frac{3+4 \iota}{2+4\iota} \times \frac{2-4 \iota}{2-4 \iota}\\=\frac{6 -12 \iota+8 \iota-16 \iota^2 }{4-16 \iota^2 }\\=\frac{ 6-4 \iota+16}{4+16}\\=\frac{22-4 \iota}{20}$

5 0

3 years ago

Question 5

ladessa [460]

Answer:

C is correct

Step-by-step explanation: c is correct great job

7 0

2 years ago

An industry representative claims that 10 percent of all satellite dish owners subscribe to at least one premium movie channel.

Greeley [361]

Answer: 1) 0.6561 2) 0.0037

Step-by-step explanation:

We use Binomial distribution here , where the probability of getting x success in n trials is given by :-

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

, where p =Probability of getting success in each trial.

As per given , we have

The probability that any satellite dish owners subscribe to at least one premium movie channel. : p=0.10

Sample size : n= 4

Let x denotes the number of dish owners in the sample subscribes to at least one premium movie channel.

1) The probability that none of the dish owners in the sample subscribes to at least one premium movie channel = $P(X=0)=^4C_0(0.10)^0(1-0.10)^{4}$

$=(1)(0.90)^4=0.6561$

∴ The probability that none of the dish owners in the sample subscribes to at least one premium movie channel is 0.6561.

2) The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.

= $P(X>2)=1-P(X\leq2)\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\\\\= 1-[0.6561+^4C_1(0.10)^1(0.90)^{3}+^4C_2(0.10)^2(0.90)^{2}]\\\\=1-[0.6561+(4)(0.0729)+\dfrac{4!}{2!2!}(0.0081)]\\\\=1-[0.6561+0.2916+0.0486]\\\\=1-0.9963=0.0037$

∴ The probability that more than two dish owners in the sample subscribe to at least one premium movie channel is 0.0037.

8 0

3 years ago

A meteorologist reported that in the month of April there were 3 inches more rainfall than normal. Write an integer to represent

BartSMP [9]

Answer:

Step-by-step explanation:

8 0

3 years ago

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