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3 years ago

Hi, thx for helping

Mathematics

1 answer:

Sliva [168]3 years ago

7 0

Answer:

C. 8.52624...

Step-by-step explanation:

The reason that C is different is because all of the other decimals are terminating decimals (meaning that they end or stop), while C is a repeating decimal, because it never ends.

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Tommy goes swimming every 12 days and hiking every 5 days. He did both today, how many days from now will he do both again? How

Alchen [17]

LCM (5,12) = 60

Roomy will do both again in 60 days ( 2 months )

3 0

2 years ago

What is the perimeter of a square with side length 5 4/5? As a mixed number.

mezya [45]

29/5+29/5+29/5+29/5= 23 1/5

8 0

3 years ago

Read 2 more answers

Suppose j varies jointly with g and v, j=1 when g=6 and v=3. Find j when g=8 and v=11

Lady bird [3.3K]

j=4.88 when g=8 and v=11

Further explanation:

When the increase/decrease in one quantity cause increase/decrease in other quantity, it is called direct variation.

Variation is always accompanied by a variation constant.

<u>Given</u>

g and v vary directly with j

IT can be written as:

j∝gv

Putting the variation constant k

j = kgv

Putting g = 6 and v=3

$j = kgv\\1 = k * 6 * 3\\1 = 18k\\k = \frac{1}{18}$

So the value of k is 1/18 which makes the equation

$j = \frac{1}{18}gv\\ Putting\ g=8\ and\ v=11\\j = \frac{1}{18} * 8 * 11\\j = \frac{88}{18}\\ j =4.88$

So, j=4.88 when g=8 and v=11

Keywords: Variation, Direct Variation

Learn more about variation at:

#LearnwithBrainly

8 0

3 years ago

Georgianna claims that in a small city renowned for its music school, the average child takes at least 5 years of piano lessons.

Allisa [31]

Answer:

The condition are

The Null hypothesis is $H_o : \mu = 5$

The Alternative hypothesis is $H_a : \mu < 5$

The check revealed that

There is sufficient evidence to support the claim that in a small city renowned for its music school, the average child takes at least 5 years of piano lessons

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 5 \ year$

The sample size is n = 20

The sample mean is $\= x = 4.6 \ years$

The standard deviation is $\sigma = 2.2 \ years$

The Null hypothesis is $H_o : \mu = 5$

The Alternative hypothesis is $H_a : \mu < 5$

So i will be making use of $\alpha = 0.05$ level of significance to test this claim

The critical value of $\alpha$ from the normal distribution table is $Z_\alpha = 1.645$

Generally the test statistics is mathematically evaluated as

$t = \frac{\= x - \mu}{ \frac{\sigma }{\sqrt{n} } }$

substituting values

$t = \frac{ 4.6 - 5}{ \frac{2.2}{\sqrt{20} } }$

$t = -0.8131$

Looking at the value of t and $Z_{\alpha }$ we see that $t < Z_{\alpha }$ so we fail to reject the null hypothesis

This implies that there is sufficient evidence to support the claim that in a small city renowned for its music school, the average child takes at least 5 years of piano lessons.

4 0

3 years ago

Which of the following is the solution to |x|-5<= -13?

kow [346]

Answer:

drishyam 2 train to bysan

Step-by-step explanation:

7 0

3 years ago

Hi, thx for helping​

Hi, thx for helping