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3 years ago

Evaluate the expression: -3.7 – 12 + (-9.49)

Mathematics

1 answer:

svp [43]3 years ago

3 0

Answer: -25.19

Step-by-step explanation:

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Rene is going to the lake to visit some friends. If the lake is 60 miles away, and Rene is driving at 40 miles per hour the enti

Tpy6a [65]

One hour and twenty minutes????

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3 years ago

PLEASE HELP ME ASAP :(

nexus9112 [7]

Step-by-step explanation:

y=4/3x+4

y int =4

x int = -1

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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.

8 0

3 years ago

A. Slope is 3.5, and (-3, 2) is on the line.

Shkiper50 [21]

Answer:

b = 12.5

Step-by-step explanation:

2=3.5(-3)+b

2=-10.5+b (you add 10.5 on both sides)

b= 12.5

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3 years ago

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yulyashka [42]

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