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3 years ago

Gina tracks the low temperatures for a city over six days. Identify the days colder than -1.75°F.

Mathematics

1 answer:

jeka57 [31]3 years ago

7 0

Answer:

Its Monday and Tuesday

Step-by-step explanation:

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The Rocky Mountain News (January 24, 1994) indicated that the 20-year mean snowfall in the Denver/Boulder region is 28.76 inches

ycow [4]

Answer:

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

Step-by-step explanation:

20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.

At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So

$H_0: \mu = 28.76$

At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So

$H_1: \mu > 28.76$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

28.76 is tested at the null hypothesis:

This means that $\mu = 28.76$

Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.

This means that $\sigma = 7.5, X = 33, n = 32$ .

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}$

$z = 3.2$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level $\alpha = 0.05$ .

Looking at the z-table, z = 3.2 has a p-value of 0.9993.

1 - 0.9993 = 0.0007

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

5 0

2 years ago

Match them together to correct answer

Verizon [17]

ANSWER

Frequency=2

Midline =3

Amplitude =5

Period =π

EXPLANATION

We have:

$f(x) = 5 \cos(x) + 3$

The given function is of the form;

$f(x) = a\cos(bx) + c$

where

a=|5| =5 is the amplitude

and b=2 is the frequency.

and 2π/b=2π/2=π

The range of the given function is

8≤y≤-2

The midline is

$\frac{8 + - 2}{2} = \frac{6}{2} = 3$

6 0

3 years ago

Read 2 more answers

The tree in johns backyard is 7 meters high. How high is it in in centimeters?

GenaCL600 [577]

Answer:

700 meters

Step-by-step explanation:

multiply the length value by 10

8 0

2 years ago

Read 2 more answers

Unit 7 polygons & quadrilaterals homework 3: rectangles Gina Wilson answer key

Irina-Kira [14]

Answer:

In the image attached you can find the Unit 7 homework.

We need to findt he missing measures of each figure.

Notice that the first figure is a rectangle, which means opposite sides are congruent so,

VY = 19

WX = 19

YX = 31

VW = 31

To find the diagonals we need to use Pythagorean's Theorem, where the diagonals are hypothenuses.

$VX^{2}=19^{2}+31^{2}\\ VX=\sqrt{361+961}=\sqrt{1322} \\VX \approx 36.36$

Also, $YW \approx 36.36$ , beacuse rectangles have congruent diagonals, which intercect equally.

That means, $ZX = \frac{VX}{2} \approx \frac{36.36}{2}\approx 18.18$

Figure number two is also a rectangle.

If GH = 14, that means diagonal GE = 28, because diagonals intersect in equal parts.

Now, GF = 11, because rectangles have opposite sides congruent.

DF = 28, because in a reactangle, diagonals are congruent.

HF = 14, because its half of a diagonal.

To find side DG, we need to use Pythagorean's Theorem, where GE is hypothenuse

$GE ^{2}=11^{2}+DG^{2}\\28^{2}-11^{2}=DG^{2}\\DG=\sqrt{784-121}=\sqrt{663}\\ DG \approx 25.75$

This figure is also a rectangle, which means all four interior angles are right, that is, equal to 90°, which means angle 11 and the 59° angle are complementary, so

$\angle 11 +59\°=90\°\\\angle 11=90\°-59\°\\\angle 11=31\°$

Now, angles 11 and 4 are alternate interior angles which are congruent, because a rectangle has opposite congruent and parallel sides.

$\angle 4 = 31\°$

Which means $\angle 3 = 59\°$ , beacuse it's the complement for angle 4.

Now, $\angle 6 = 59\°$ , because it's a base angle of a isosceles triangle. Remember that in a rectangle, diagonals are congruent, and they intersect equally, which creates isosceles triangles.