What is the anwser for 4n-2n=4

The seats at a local baseball stadium are arranged so that each row has five more seats than the row below it. If there are four

brilliants [131]

Answer:

Thus the last row has 119 seats.

The total number of seats in 24 rows = 1476

Step-by-step explanation:

The number of seats in each row make an arithmetic series. We will use arithmetic equation to find the number of seats in last row:

An = a1+ (n-1)d

An = 4+(24-1)5

An = 4 + (23)(5)

An = 4 + 115

An = 119

Thus the last row has 119 seats.

Now to find the sum of seats we will apply the formula:

Sn = n(a1 + an)/2

Sn = 24(4+119)/2

Sn = 24(123) /2

Sn = 1476 .....

The total number of seats in 24 rows = 1476....

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3 years ago

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A gumball machine contains 23 green gumballs, 52 red gumballs, 34 blue gumballs, 61 yellow gumballs, and 30 pink gumballs. What

Veseljchak [2.6K]

52% of 200
52/100=26/50=13/25•200/1= 2600/25=2600 divided by 25 is 104
The answer is 104

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3 years ago

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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%