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3 years ago

HCF and LCM for 4 8 15and example

Mathematics

2 answers:

Romashka [77]3 years ago

5 0

Answer:

1 and 120

Step-by-step explanation:

kolbaska11 [484]3 years ago

3 0

Answer:

HCF of 4 8 15 = 1

LCM of 4 8 15 = 120

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Ann [662]

Answer:

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Step-by-step explanation:

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3 years ago

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Use the formula for compound interest to find A.

Crank

Easy peasy
A=20,000(1+0.13)^2
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3 years ago

Solve the equation 5 m + 4 m equals 72

gulaghasi [49]

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Step-by-step explanation:

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A ribbon is 40 inches long. We want to cut the ribbon into pieces. Each piece will be of an inch. How many pieces will we have?

mafiozo [28]

Answer:

Step-by-step explanation:

What we need here is to know how many of 1 inch are there in 40 inches

This is straightforward, what we need to do here is to imagine having to mark 1 inch length on the ribbon using a mark of a scissors.

What we will notice is that we need to make 40 marks to have exhausted the entire length of the ribbon.

This is obtained by dividing the total 40 by 1 which is 40

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3 years ago

A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12. Suppose that we take a random

Zielflug [23.3K]

Answer: The 95% confidence interval for the mean of x is (94.08, 101.92) .

Step-by-step explanation:

We are given that ,

A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.

i.e. $\sigma= 12$

Also, it is given that , Sample mean $\overline{x}=98$ having sample size : n= 36

For 95% confidence ,

Significance level : $\alpha=1-0.95=0.05$

By using the z-value table , the two-tailed critical value for 95% Confidence interval :

$z_{\alpha/2}=z_{0.025}=1.96$

We know that the confidence interval for unknown population mean $(\mu)$ is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

, where $\overline{x}$ = Sample mean

$\sigma$ = Population standard deviation

$z_{\alpha/2}$ = Critical z-value.

Substitute all the given values, then the required confidence interval will be :

$98\pm (1.96)\dfrac{12}{\sqrt{36}}$

$=98\pm (1.96)\dfrac{12}{6}$

$=98\pm (1.96)(2)$

$=98\pm 3.92=(98-3.92,\ 98+3.92)\\\\=( 94.08,\ 101.92)$

Therefore, the 95% confidence interval for the mean of x is (94.08, 101.92) .

6 0

3 years ago