Find the value of x. Round the answer to the nearest tenth, if needed.

In a game of tug of war, your team changes -2 1/2 feet in position every 10 seconds. What is your change in position after 40 se

Airida [17]

The answer is -10

-2 1/2 x 4 = -10

8 0

3 years ago

on a certain hot Summer's Day 613 people use the public swimming pool the daily prices are $1.75 for children $2.25 for adults t

Elodia [21]

389 children and 224 adults were at the pool that day.

Step-by-step explanation:

Cost of one child ticket = $1.75

Cost of one adult ticket = $2.25

Total people = 613

Total receipts worth = $1184.75

Let,

Number of children = x

Number of adults = y

According to given statement;

x+y=613 Eqn 1

1.75x+2.25y=1184.75 Eqn 2

Multiplying Eqn 1 by 1.75

$1.75(x+y=613)\\1.75x+1.75y=1072.75\ \ \ Eqn\ 3$

Subtracting Eqn 3 from Eqn 2

$(1.75x+2.25y)-(1.75x+1.75y)=1184.75-1072.75\\1.75x+2.25y-1.75x-1.75y=112.00\\0.50y=112.00$

Dividing both sides by 0.50

$\frac{0.50y}{0.50}=\frac{112.00}{0.50}\\y=224$

Putting y=224 in Eqn 1

$x+224=613\\x=613-224\\x=389\\$

389 children and 224 adults were at the pool that day.

Keywords: linear equation, subtraction

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6 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%