What’s the answer pls

Ann [662]

The roller coaster is 150 ft

8 0

3 years ago

The following data represent the pH of rain for a random sample of 12 rain dates in Tucker County, West Virginia. A normal proba

Sphinxa [80]

Answer:

Step-by-step explanation:

n = 12

Mean = (4.58 + 5.72 + 4.77 + 4.76 + 5.19 + 5.05 + 4.80 + 4.77 + 4.75 + 5.02 + 4.74 + 4.56)/12 = 4.8925

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (4.58 - 4.8925)^2 + (5.72 - 4.8925)^2 + (4.77 - 4.8925)^2 + (4.76 - 4.8925)^2 + (5.19 - 4.8925)^2 + (5.05 - 4.8925)^2 + (4.80 - 4.8925)^2 + (4.77 - 4.8925)^2 + (4.75 - 4.8925)^2 + (5.02 - 4.8925)^2 + (4.74 - 4.8925)^2 + (4.56 - 4.8925)^2 = 1.122225

Standard deviation = √(1.122225/12

s = 0.31

a) Point estimate = sample mean = 4.8925

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

Margin of error = z × s/√n

Where

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 12 - 1 = 11

b) Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 2.201

Margin of error = 2.201 × 0.31/√12

= 0.197

95% confidence interval = 4.8925 ± 0.197

Upper limit = 4.8925 + 0.197 = 5.0895

Lower limit = 4.8925 - 0.197 = 4.6955

We are 95% confident that the population mean of the rain water ph lies between 4.6955 and 5.0895

c) For 99% confidence level, z = 3.106

Margin of error = 3.106 × 0.31/√12

= 0.278

99% confidence interval = 4.8925 ± 0.278

Upper limit = 4.8925 + 0.278 = 5.1705

Lower limit = 4.8925 - 0.278 = 4.6145

We are 99% confident that the population mean of the rain water ph lies between 4.6145 and 5.1705

d) The interval gets wider as the confidence level is increased. This is logical since the test score is higher for 99% and therefore, increases the range of values. Since we want to be more confident, the range of values must be extended.

3 0

4 years ago

A bank offers a APR of 3.1 compound daily

Airida [17]

Sure does -----------------

8 0

4 years ago

PLEASE HELP

sweet [91]

To model and solve our situation we are going to use the equation: $s= \frac{d}{t}$
where
$s$ is speed
$d$ is distance
$t$ is time

1. We know that the distance between the cities is 2400 miles, so $d=2400$ . We also know that the speed of the plane is 450 mi/h. Since we don't know the speed of the air, $S_{a}=?$ . We don't know how much the westward trip takes, so $t_{w}=?$ , and we also don't know how much the eastward trip takes, so $t_{e}=?$ .

Going westward. Here the plane is flying against the air, so we need to subtract the speed of the air from the speed of the plane:
$450-S_{a}= \frac{2400}{t_{w} }$
Going eastward. Here the plane is flying with the the air, so we need to add the speed of the air to the speed of the plane:
$450+S_{a}= \frac{2400}{t_{e} }$

2. We know for our problem that the round trip takes 11 hours; so the total time of the trip is 11, $t_{t}=11$ . Notice that we also know that the total time of the trip equals time of the tip going westward plus time of the trip going eastward, so $t_{t}=t_{w}+t_{e}$ . Since we know that the total trip takes 11 hours, we can replace that value in our total time equation and solve for $t_{w}$ :
$11=t_{w}+t_{e}$
$t_{w}=11-t_{e}$

Now we can replace $t_{w}$ in our going westward equation to model our round trip with a system of equations:
$450-S_{a}= \frac{2400}{t_{w}}$
$450-S_{a}= \frac{2400}{11-t_{e} }$ equation (1)
$450+S_{a}= \frac{2400}{t_{e}}$ equation (2)

3. To solve our system of equations, we are going to solve for $t_{e}$ in equations (1) (2):

From equation (1)
$450-S_{a}= \frac{2400}{11-t_{e} }$
$11-t_{e}= \frac{2400}{450-S_{a} }$
$-t_{e}= \frac{2400}{450-S_{a} } -11$
$t_{e}=11- \frac{2400}{450-S_{a} }$
$t_{e}= \frac{4950-11S_{a} -2400}{450-S_{a} }$
$t_{e}= \frac{2550-11S_{a} }{450-S_{a} }$ equation (3)

From equation (2):
$450+S_{a}= \frac{2400}{t_{e} }$
$t_{e}= \frac{2400}{450+S_{a} }$ equation (4)

Replacing (4) in (3)
$\frac{2400}{450+S_{a}} = \frac{2550-11S_{a}}{450-S_{a} }$
Now, we can solve for $S_{a}$ to find the speed of the wind:
$2400(450-S_{a})=(450+S_{a})(2550-11S_{a})$
$1080000-2400S_{a}=1147500-4950S_{a}+2550S_{a}-11(S_{a})^{2}$
$11(S_{a})^{2}-67500=0$
$11(S_{a})^{2}=67500$
$(S_{a})^{2}= \frac{67500}{11}$
$S_{a}=+/- \sqrt{ \frac{67500}{11} }$
Since speed cannot be negative, the solution of our equation is:
$S_{a}= \sqrt{ \frac{67500}{11} }$
$S_{a}=78.33$

We can conclude that the speed of the wind is 78 mph.

3 0

4 years ago

The product of a number and 5 is less than the number

KatRina [158]

Answer:

5

Step-by-step explanation:

5 0

3 years ago