A) Which inequality is shown on this graph

A plane is traveling at a rate of 500 miles per hour. The remaining fuel can last for up to 1500 miles.

Tju [1.3M]

Answer:

d. since we know it can travel 500 mph and it has enough fuel to fly for another 1500 miles. it can fly for 3 more hours, the reason D is the correct answer is since the plane can't fly more than 3 hours so it will fly for 3 hours exactly or less

6 0

3 years ago

-1/2x+3=-x+7 I just don't know how to take the fraction out :(

wariber [46]

Answer:

x = 8

Step-by-step explanation:

$- \frac{1}{2} x + 3 = - x + 7$

Collect like terms and simplify

$- \frac{1}{2} x + x = 7 - 3 \\ \frac{1}{2} x = 4$

Cross Multiply

$x \times 1 = 2 \times 4 \\ x = 8$

3 0

4 years ago

Read 2 more answers

“Please help!!” **I WILL GIVE BRAINLIEST!** -MUST EXPLAIN ANSWER-

seropon [69]

i can't see the picture

5 0

3 years ago

Como se divide 0.75÷5 con cociente menor que 1 ayudaaaaaa

REY [17]

La respuesta correcta es .15

7 0

3 years ago

A publisher wants to estimate the mean length of time (in minutes) all adults spend reading newspapers. To determine this estima

liq [111]

Answer:

a) $\bar X=9.07$

b) The 95% confidence interval is given by (8.197;9.943)

c) $m=2.14 \frac{1.580}{\sqrt{15}}=0.873$

d) 3 possible ways

1) Increasing the sample size n.

2) Reducing the variability. If we have more data probably we will have less variation.

3) Lower the confidence level. Because if we have lower confidence then the quantile from the t distribution would belower and tthe margin of error too.

Step-by-step explanation:

Notation and definitions

n=15 represent the sample size

$\bar X= 9.07$ represent the sample mean

$s=1.580$ represent the sample standard deviation

m represent the margin of error

Confidence =95% or 0.95

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".

Part a: Find the point estimate of the population mean.

The point of estimate for the population mean $\mu$ is given by:

$\bar X =\frac{\sum_{i=1}^{n} x_i}{n}$

The mean obteained after add all the data and divide by 15 is $\bar X=9.07$

Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

$df=n-1=15-1=14$

We can find the critical values in excel using the following formulas:

"=T.INV(0.025,14)" for $t_{\alpha/2}=-2.14$

"=T.INV(1-0.025,14)" for $t_{1-\alpha/2}=2.14$

The critical value $tc=\pm 2.14$

Part c: Calculate the margin of error (m)

First we need to calculate the standard deviation given by this formula:

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$

s=1.580

The margin of error for the sample mean is given by this formula:

$m=t_c \frac{s}{\sqrt{n}}$

$m=2.14 \frac{1.580}{\sqrt{15}}=0.873$

Part b: Calculate the confidence interval

The interval for the mean is given by this formula:

$\bar X \pm t_{c} \frac{s}{\sqrt{n}}$

And calculating the limits we got:

$9.07 - 2.14 \frac{1.580}{\sqrt{15}}=8.197$

$9.07 + 2.14 \frac{1.580}{\sqrt{15}}=9.943$

The 95% confidence interval is given by (8.197;9.943)

Part d: How can we reduce the margin of error?

We can reduce the margin of error on the following ways:

1) Increasing the sample size n.

2) Reducing the variability. If we have more data probably we will have less variation.

3) Lower the confidence level. Because if we have lower confidence then the quantile from the t distribution would belower and tthe margin of error too.

3 0

3 years ago