Find decimal notation. 66 7/8%

Explain why the exponents cannot be added in the product 12^3 x 11^3.

sergij07 [2.7K]

Answer:

see explanation

Step-by-step explanation:

The exponents can only be added if the bases are the same

Here one base is 12 and the other is 11

So the rule of addition is not applicable.

4 0

2 years ago

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Please help! its due in 10 mins

den301095 [7]

Answer:

144

Step-by-step explanation:

sorry if im wrong its been awhile since ive done this

4 0

3 years ago

Mrs. Kennedy is buying pencils for each of 315 students. The pencils are sold in boxed of tens. How can she use rounding to deci

ExtremeBDS [4]

Mrs.Kennedy can divide 315 by 10 and then when she gets her answer she can round up to get her answer.

3 0

3 years ago

In an examination it is required to get 40% of the aggregate marks to pass. A student gets 261 marks and is declared failed by 4

svetoff [14.1K]

The solution to the problem is as follows:

→x− 4% (x) = 261

→x − 4x/100 = 261

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

4 0

3 years ago

A random sample of 150 men found that 88 of the men excercise regularly, while a random sample 200 women found that 130 of the w

melamori03 [73]

Answer:

The hypothesis is:

H₀: $p_{X}-p_{Y}=0$ .

Hₐ: $p_{X}-p_{Y}$ .

Step-by-step explanation:

Let X = number of men who exercise regularly and Y = number of women who exercise regularly.

The information provided is:

$n_{X}=150\\X=88\\n_{Y}=200\\Y=130$

Compute the sample proportion of men and women who exercise regularly as follows:

$\hat p_{X}=\frac{X}{n_{X}}=\frac{88}{150}=0.587$

$\hat p_{Y}=\frac{Y}{n_{Y}}=\frac{130}{200}=0.65$

The random variable X follows a Binomial distribution with parameters n = 150 and $\hat p_{X}=0.587$ .

The random variable Y also follows a Binomial distribution with parameters n = 200 and $\hat p_{Y}=0.65$ .

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\hat p=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

So, the sampling distribution of the proportion of men and women who exercise regularly follows a Normal distribution.

A two proportion z-test cab be performed to determine whether the proportion of women is more than men who exercise regularly.

The hypothesis for this test cab be defined as:

H₀: The proportion of women is same as men who exercise regularly, i.e. $p_{X}-p_{Y}=0$ .

Hₐ: The proportion of women is more than men who exercise regularly, i.e. $p_{X}-p_{Y}$ .

6 0

3 years ago

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