forgets how to do this

Mathematics

2 answers:

hammer [34]3 years ago

8 0

Answer:

-12

Step-by-step explanation:

-8-(-12= -8+12

well, 12-8=4, so -8+12=4

Aleksandr-060686 [28]3 years ago

5 0

Answer:

Step-by-step explanation:

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7 mark question. The answers on brainly only got me 3 so please help.

Nady [450]

Answer:

See step by step answer

Step-by-step explanation:

the factors are (x+22)(x-10)=0

so x = -22 and 10, but a dimension cannot be negative, so the answer for x is 10 and then if you plug it back in the dimensions are 4 cm, 11 cm and 21 cm.

When you multiply 4 * 11 * 21 = 924, so you know it's correct.

8 0

3 years ago

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:

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

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