HELP dont answer this

Rewrite the expression in the form 6^n 6^-6/6^-5

oee [108]

Answer:

6^-1

Step-by-step explanation:

6^-6 - (-5) = 6^(-6 + 5) = 6^(-1)

A lot of things are going on in this question. Notice how the 5 is handled. It starts off in the denominator as - 5 and when brought to the numerator it becomes -(-5) which makes it + 5.

The answer is then easy. It is 6^(-6 + 5) = 6^-1

You could give the answer as 1/6 but that is not what the question says.

8 0

3 years ago

Joseph and Hanna are 250 feet apart when they start walking toward one another. They are walking at the same speed, so whenever

Evgen [1.6K]

I dont know the answer

4 0

3 years ago

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%

It means the the level of significance $\alpha$ is:

$\alpha =1-0.98=0.02$

4 0

3 years ago

Mr. warren has a bag of nails assorted lengths. he asks his sons paul and tim to sort them by length. paul pours 3/4 of the bag

8090 [49]

Answer:

Step-by-step explanation:

he has 1/4 remaining as he has only used 3/4

5 0

2 years ago

18.

ASHA 777 [7]

Answer:

19/100 which is already simplified( it can't be simplified)

How I got the answer: subtract 81 from 100 and you'll get 19.

Put 19 over 100 (19/100) and there's the answer.

3 0

3 years ago