I need help how many negative integer chips did you cross out. From-9-(-4)

At a regional soccer tournament, tickets for two adults and seven students cost $60. Tickets for three adults and eleven stude

den301095 [7]

Answer:

The price of an adult ticket is $9 and the price of a student ticket is $6

Step-by-step explanation:

Create a system of equations where x is the price of an adult ticket and y is the price of a student ticket:

2x + 7y = 60

3x + 11y = 93

Solve by elimination by multiplying the top equation by 3 and the bottom equation by -2, to cancel out the x terms:

6x + 21y = 180

-6x - 22y = -186

Add them together:

-y = -6

y = 6

Then, plug in 6 as y into one of the equations to solve for x:

2x + 7y = 60

2x + 7(6) = 60

2x + 42 = 60

2x = 18

x = 9

So, the price of an adult ticket is $9 and the price of a student ticket is $6

8 0

3 years ago

What is the significance of a correlation coefficient of 0.88?

AleksandrR [38]

The maximum positive value that the correlation coefficient can have is 1.
Thus, 0.88 is pretty close, indicating a positive slope for the regression line and points lying close to that line.

6 0

3 years ago

But how is is it 1 hour and 45 minutes(someone please respond this is due tomorrow)I need work

astraxan [27]

Because it is every hour there's 60min so that's and hour the the next 45 min are different

4 0

3 years ago

A study was conducted and two types of engines, A and B, were compared. Fifty experiments were performed using engine A and 75 u

USPshnik [31]

Answer:

a) -6 mpg.

b) 2.77 mpg

c) The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem, and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Gas mileage A: Mean 36, standard deviation 6, sample of 50:

So

$\mu_A = 36, s_A = \frac{6}{\sqrt{50}} = 0.8485$

Gas mileage B: Mean 42, standard deviation 8, sample of 50:

So

$\mu_B = 42, s_B = \frac{8}{\sqrt{50}} = 1.1314$

Distribution of the difference:

Mean:

$\mu = \mu_A - \mu_B = 36 - 42 = -6$

Standard error:

$s = \sqrt{s_A^2+s_B^2} = \sqrt{0.8485^2+1.1314^2} = 1.4142$

A. Find the point estimate.

This is the difference of means, that is, -6 mpg.

B. Find the margin of error

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = zs = 1.96*1.4142 = 2.77$

The margin of error is of 2.77 mpg

C. Construct the 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results(5 pts)

The lower end of the interval is the sample mean subtracted by M. So it is -6 - 2.77 = -8.77 mpg

The upper end of the interval is the sample mean added to M. So it is -6 + 2.77 = -3.23 mpg

The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

8 0

3 years ago

An experimenter conducted a two-tailed hypothesis test on a set of data and obtained a p-value of 0.44. If the experimenter had

Lyrx [107]

Answer:

Correct option is (C).

The possible value of the p-value for a one-tailed test are 0.22 and 0.78.

Step-by-step explanation:

The p-value is the probability of acquiring a result as extreme as the observed result, assuming the null hypothesis statement is true.

The p value of a test is:

Left-tailed test: $P(TS$