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ehidna [41]
3 years ago
5

F(x) =5x^2-20x+3 how to find minimum

Mathematics
1 answer:
Leya [2.2K]3 years ago
3 0

Answer:

(2,-17) should be the minimum.

Step-by-step explanation:

The minimum of a quadratic function occurs at x=-\frac{b}{2a} . If a is positive, the minimum value of the function is f(-\frac{b}{2a})

f_{min}x=ax^2+bx+c occurs at x=-\frac{b}{2a}

Find the value of x=-\frac{b}{2a}

x = 2

evaluate f(2).

replace the variable x with 2 in the expression.

f(2)=5(2)^2-20(2)+3

simplify the result.

f(2)=5(4)-20(2)+3

f(2)=20-40+3

f(2)=-17

The final answer is -17

Use the x and y values to find where the minimum occurs.

HOPE THIS HELPS!

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all of the calandars at the bookstore are on sale this weekend for $5.75. if joanne bought $149.50 worth of calanders this weeke
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A manager of a grocery store wants to determine if consumers are spending more than the national average. The national average i
strojnjashka [21]

The valid conclusions for the manager based on the considered test is given by: Option

<h3>When do we perform one sample z-test?</h3>

One sample z-test is performed if the sample size is large enough (n  > 30) and we want to know if the sample comes from the specific population.

For this case, we're specified that:

  • Population mean = \mu = $150
  • Population standard deviation = \sigma = $30.20
  • Sample mean = \overline{x} = $160
  • Sample size = n = 40 > 30
  • Level of significance = \alpha = 2.5% = 0.025
  • We want to determine if the average customer spends more in his store than the national average.

Forming hypotheses:

  • Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get: H_0: \mu_0 \leq \mu = 150
  • Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically H_1: \mu_0 > \mu = 150

where \mu_0 is the hypothesized population mean of the money his customer spends in his store.

The z-test statistic we get is:

z = \dfrac{\overline{x} - \mu_0}{\sigma/\sqrt{n}} = \dfrac{160 - 150}{30.20/\sqrt{40}} \approx 2.094

The test is single tailed, (right tailed).

The critical value of z at level of significance 0.025 is 1.96

Since we've got 2.904 > 1.96, so we reject the null hypothesis.

(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).

Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.

Learn more about one-sample z-test here:

brainly.com/question/21477856

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