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3 years ago

PLEASE SHOW WORK!! THANK YOU

Mathematics

2 answers:

erik [133]3 years ago

6 0

Answer:

[See Below]

Step-by-step explanation:

✦ So it must add up to 360dg so add:

✧ 115dg + 85dg + 60dg = 260dg

✦ Now subtract:

✧ 360dg - 260dg = 100dg

So X would be 100 degrees.

~<em>Hope this helps Mate. If you need anything feel free to message me. </em>

bearhunter [10]3 years ago

4 0

You can use Cymath it shows the work for you

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FREEEEEEEEEEEEEEEEEEEEEEEE POINTS!!!!!!!!!!!!!

muminat

Bless ur kind soul :o

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3 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.

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3 years ago