Find the gradient of the line between each pair of points A(2,2) and B(6,6)

Which correctly uses bar notation to represent the repeating decimal

ioda

Answer:

2nd one..

Step-by-step explanation:

but although all are wrongmm barnotation is not used

6 0

3 years ago

The figure shown is a rhombus.Its measurements are in cm.Calculate

tia_tia [17]

I need a picture of the figure so I can calculate it.

3 0

2 years ago

At the theater, the Garcia family bought 2 adult tickets and 3 children's tickets for $44.50. The Smith family bought 3 adult ti

kondaur [170]

Answer:

Adult= $11

Children = $7.5

Step-by-step explanation:

Let x represent adult ticket and y represent children ticket

2x + 3y= 44.50........equation 1

3x + 6y= 78........equation 2

From equation 1

2x + 3y= 44.50

2x= 44.50-3y

x= 44.50-3y/2

Substitute 44.50-3y/2 for x in equation 2

3x+ 6y= 78

3(44.50-3y/2) + 6y= 78

66.75- 4.5y +6y= 78

66.75 + 1.5y= 78

1.5y= 78-66.75

1.5y= 11.25

y= 11.25/1.5

y = 7.5

Substitute 7.5 for y in equation 1

2x + 3y = 44.50

2x + 3(7.5)= 44.50

2x + 22.5= 44.50

2x = 44.50-22.5

2x= 22

x= 22/2

x= 11

Hence the price of adult ticket is $11 and the price of children ticket is $7.5

6 0

2 years ago

Read 2 more answers

Find the variance and standard deviation of the data. Round your answers to the

LenaWriter [7]

83, 85, 89, 92, 95 it should be this

6 0

2 years ago

Suppose 60% of American adults believe Martha Stewart is guilty of obstruction of justice and fraud related to insider trading.

Ierofanga [76]

Answer:

Since the sample size is large enough (n>30) and the probability of success is near to 0.5, and we have that $n\hat p = 60>10$ and $n(1-\hat p) = 40>10$ we can assume that the distribution for $\hat p$ is normal

The population proportion have the following distribution

$\hat p \sim N(\hat p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The mean is given by:

$\mu_p = 0.6$

$\sigma_p = \sqrt{\frac{0.6*(1-0.6)}{100}}=0.04899$

So then we can conclude that the best answer would be:

c. approximately Normal, with mean 0.6 and standard error 0.04899

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

Solution to the problem

Since the sample size is large enough (n>30) and the probability of success is near to 0.5, and we have that $n\hat p = 60>10$ and $n(1-\hat p) = 40>10$ we can assume that the distribution for $\hat p$ is normal

The population proportion have the following distribution

$\hat p \sim N(\hat p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The mean is given by:

$\mu_p = 0.6$

$\sigma_p = \sqrt{\frac{0.6*(1-0.6)}{100}}=0.04899$

So then we can conclude that the best answer would be:

c. approximately Normal, with mean 0.6 and standard error 0.04899

8 0

3 years ago