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3 years ago

What is the range of h (x)

1 answer:

6 0

The range of the function is the set of all possible outputs, that is, the set of all values obtained by applying the function to elements of the domain. So the set of all values which can be obtained by applying h(x) to an element of its domain is {−4,0,5,60} , and thus that is the range of h(x) .

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Using the digits 0 to 9 at most one Time each,fill in the boxes below to write equations whose solution is 1/2

andrezito [222]

Answer:

The equation is 4x + 5 = 2x + 6

Step-by-step explanation:

The solution is $\frac{1}{2}$ means x = $\frac{1}{2}$

Let us put the coefficient of x in the left hand side = 4 and the coefficient of x in the right hand side = 2

∵ 4x - 2x = 2x

Now we need two numerical terms their difference is 1 like 5 and 6 or 6 and 7 or 7 and 8 or 8 and 9

Let us take 5 and 6

∵ 4x + 5 = 2x + 6

- Subtract 2x from both sides

∴ 2x + 5 = 6

- Subtract 5 from both sides

∴ 2x = 1

- Divide both sides by 2

∴ x = $\frac{1}{2}$

∴ The solution is $\frac{1}{2}$

The equation is 4x + 5 = 2x + 6

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3 years ago

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Mila [183]

Answer:

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Step-by-step explanation:

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Apple cost $0.89 pound how much will 25 pounds cost

AfilCa [17]

22.25

multiply them by each other and you will get your answer 22.25

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Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - α)% confidence interval for the population variance is given as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

It is provided that:

n = 20

s = 3.9

Confidence level = 95%

⇒ α = 0.05

Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$