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irakobra [83]
3 years ago
5

A statistics student records the servings of vegetables Natalia and Zane eat each weekday. The table shows the results.

Mathematics
1 answer:
Softa [21]3 years ago
3 0

Answer:

The Answer is D)

Step-by-step explanation:

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The Miller family drove 1,240 miles on their road trip. If their trip took 4 days, how many miles did they travel per day?
nikitadnepr [17]

Answer:  310

Step-by-step explanation: 1240 miles /4 days = 310 miles per one day

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(38x+61y)-(3x-11y) simplify this expression
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▹ Answer

<em>35x + 72y</em>

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▹ Step-by-Step Explanation

(38x + 61y) - (3x - 11y)

38x + 61y - (3x - 11y)

38x + 61 y - 3x + 11y

35x + 61y + 11y

35x + 72y

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4 0
3 years ago
Having a problem solving for h in this algebra problem. A=1/2h(B+b); A=35, B=6, b=1 <br> So h=?
ivann1987 [24]
35 = 1/2h (6+1)
35 = 1/2 h + 7
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2x+6y=-24 converted into a slope intercept?
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Step-by-step explanation:

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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
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