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Triss [41]
3 years ago
10

11) Find the value for x for both diagrams please help me

Mathematics
1 answer:
inysia [295]3 years ago
3 0

Answer:

Step-by-step explanation:

11 a).

20x + 5 + 24x - 1 = 180

44x = 176

x = 176 / 44

x = 4

b). 6x = 5x + 10

x = 10

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3 choice

Step-by-step explanation:

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What is the square root of ten
lyudmila [28]
10 is not a perfect square I believe
8 0
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Suppose that a stock is currently selling for $100. The change in the stock's price during the next year follows a normal random
Basile [38]

Answer:

The probability that the stock will sell for $85 or less in a year's time is 0.10.

Step-by-step explanation:

Let <em>X</em> = stock's price during the next year.

The random variable <em>X</em> follows a normal distribution with mean, <em>μ</em> = $100 + $10 = $110 and standard deviation, <em>σ</em> = $20.

To compute the probability of a normally distributed random variable we first need to compute the <em>z</em>-score for the given value of the random variable.

The formula to compute the <em>z</em>-score is:

z=\frac{X-\mu}{\sigma}

Compute the probability that the stock will sell for $85 or less in a year's time as follows:

Apply continuity correction:

P (X ≤ 85) = P (X < 85 - 0.50)

                = P (X < 84.50)

                =P(\frac{X-mu}{\sigma}

                =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the stock will sell for $85 or less in a year's time is 0.10.

6 0
3 years ago
Product of the polynomial (2y+3)(3y^2+4y+5)
OleMash [197]
The answer is 6y^3+17y^2+22y+15
5 0
3 years ago
Some parts of California are particularly earthquake- prone. Suppose that in one metropolitan area, 25% of all homeowners are in
aleksandrvk [35]

Answer:

a. Binomial random variable (n=4, p=0.25)

b. Attached.

c. X=1

Step-by-step explanation:

This can be modeled as a binomial random variable, with parameters n=4 (size of the sample) and p=0.25 (proportion of homeowners that are insured against earthquake damage).

a. The probability that X=k homeowners, from the sample of 4, have eartquake insurance is:

P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{4}{k} 0.25^{0}\cdot0.75^{4}

The sample space for X is {0,1,2,3,4}

The associated probabilties are:

P(x=0) = \dbinom{4}{0} p^{0}(1-p)^{4}=1*1*0.3164=0.3164\\\\\\P(x=1) = \dbinom{4}{1} p^{1}(1-p)^{3}=4*0.25*0.4219=0.4219\\\\\\P(x=2) = \dbinom{4}{2} p^{2}(1-p)^{2}=6*0.0625*0.5625=0.2109\\\\\\P(x=3) = \dbinom{4}{3} p^{3}(1-p)^{1}=4*0.0156*0.75=0.0469\\\\\\P(x=4) = \dbinom{4}{4} p^{4}(1-p)^{0}=1*0.0039*1=0.0039\\\\\\

b. The histogram is attached.

c. The most likely value for X is the expected value for X (E(X)).

Is calculated as:

E(X)=np=4\cdot0.25=1

6 0
3 years ago
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