What is 145 ÷ 891 + 153 ÷ 141 + 1/1?

Mathematics

2 answers:

34kurt2 years ago

8 0

Answer:

≈ 2.24784

Step-by-step explanation:

145÷891+153÷141+ 1/1

Any expression divided by 1 remains the same.

145÷891+153÷141+ 1

Write the division as a fraction.

145+153+1

891 141

Calculate the sum.

94133 = 2 10379 ≈ 2.24784

41877 41877

ruslelena [56]2 years ago

7 0

Answer: Exact form: 94133 / 41877. Mixed number form: 2 10379 / 41877

Step-by-step explanation: divide 145 and 891. Then add 153. Then divide again 141 and 153. Then add 1/1.

Hope this helps!

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L=53+D What the is length of the road after 22 days ?

Savatey [412]

<h3>After 22 days the length of the road is = 75 units.</h3>

Step-by-step explanation:

Given,

L=53 +D L = the length of road and D = number of day

After 22 days the length of the road is = (53+ 22)

= 75 units

4 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:

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

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