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3 years ago

Pls help!! if f(x) = 2x^2-4x and g(x)=5-2x, evaluate f(x) - g(x) for x=5

Mathematics

1 answer:

Sindrei [870]3 years ago

7 0

Answer is C=35

How? Plug 5 into all the x's and solve for both g(x) and f(x) then finish off by subtracting f(x) from g(x).

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Mike deposited $6500 into two saving accounts bearing simple interest. One of the accounts has an interest rate of 3% while the

ololo11 [35]

0.03x+0.06 (6500-x)=225
Solve for x
0.03x+390-0.06x=225
0.03x-0.06x=225-390
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3 years ago

In the Hawaiian Beach Boy surf board vendor scenario, what if the fine was increased to $190 but the probability of a fine decre

zloy xaker [14]

Answer:

$190.50

Step-by-step explanation:

Expected value is the sum of each possible income multiplied by its probability.

There's a 5% chance that the vendor makes $200 and loses $190 (net gain of $10).

There's a 95% chance that the vendor makes $200 and loses $0 (net gain of $200).

So the expected value is:

Exp(RS) = $10 × 0.05 + $200 × 0.95

Exp(RS) = $190.50

5 0

3 years ago

Suppose a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%. A cell phon

leva [86]

Answer:

We conclude that the actual percentage of households is equal to 30%.

Step-by-step explanation:

We are given that a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%.

Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

Let p = proportion of households that have three cell phones NOT known.

So, Null Hypothesis, $H_0$ : p = 30% {means that the actual percentage of households is equal to 30%}

Alternate Hypothesis, $H_A$ : p $\neq$ 30% {means that the actual percentage of households different from 30%}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of households having three cell phones = $\frac{43}{150}$ = 0.29

n = sample of households = 150

So, the test statistics = $\frac{0.29-0.30}{\sqrt{\frac{0.30(1-0.30)}{150} } }$

= -0.27

The value of z test statistic is -0.27.

Also, P-value of the test statistics is given by;

P-value = P(Z < -0.27) = 1 - P(Z $\leq$ 0.27)

= 1 - 0.6064 = 0.3936

Now, at 1% significance level the z table gives critical value of -2.58 and 2.58 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual percentage of households is equal to 30%.

4 0

3 years ago