Write doon the Multiplicative inverses<br>a) - 167942 <br>b)

The average cost of a gallon of gas during July was $2.45. The average cost of a gallon of gas during August was $2.78. What is

nekit [7.7K]

Step One

Figure out the actual change.

Change = 2.78 - 2.45

Change = 0.33

Step 2

Find the % change

% Change = 0.33/2.45 * 100 = 13.5%

4 0

4 years ago

(a) दिइएको चतुभुर्जको क्षेत्रफल पत्ता

My name is Ann [436]

Answer:

मैं वास्तव में अच्छा हूं

मैं धाराप्रवाह हिंदी बोलता हूं

5 0

3 years ago

A school system is reducing the amount of dumpster loads of trash removed each week. In week 5, there were 40 dumpster loads of

iragen [17]

Plugging in (5 weeks, 40 loads) and (10 weeks, 30 loads) into each choice indicates that f(x) = -2x + 50 is the only correct answer.

4 0

3 years ago

Read 2 more answers

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

Which is an internal environmental factor that affects heredity?

Naddik [55]

Answer: D.Hormones

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

Write doon the Multiplicative inversesa) - 167942 b) - 8​

Write doon the Multiplicative inversesa) - 167942 b) - 8