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Deffense [45]
3 years ago
14

How do we find 4and1/4 hours after 11:20pm?

Mathematics
1 answer:
tigry1 [53]3 years ago
5 0

there are 60 minutes in 1 hour, so 1/4 of an hour is 60(1/4), namely 15 minutes.


11:20pm + 4 hours, is 11+4:20, namely 15:20, of course the time system only uses up to 12, so that has to be 3:20, and then we add the 15 minutes.


11+4: 20 + 15.........3:35am.

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The scientist performs additional analyses and observes that the number of major earthquakes does appear to be decreasing but wo
kifflom [539]

Answer:

Step-by-step explanation:

Hello!

A regression model was determined in order to predict the number of earthquakes above magnitude 7.0 regarding the year.

^Y= 164.67 - 0.07Xi

Y: earthquake above magnitude 7.0

X: year

The researcher wants to test the claim that the regression is statistically significant, i.e. if the year is a good predictor of the number of earthquakes with magnitude above 7.0 If he is correct, you'd expect the slope to be different from zero: β ≠ 0, if the claim is not correct, then the slope will be equal to zero: β = 0

The hypotheses are:

H₀: β = 0

H₁: β ≠ 0

α: 0.05

The statistic for this test is a student's t: t= \frac{b - \beta }{Sb} ~~t_{n-2}

The calculated value is in the regression output t_{H_0}= -3.82

This test is two-tailed, meaning that the rejection region is divided in two and you'll reject the null hypothesis to small values of t or to high values of t, the p-value for this test will also be divided in two.

The p-value is the probability of obtaining a value as extreme as the one calculated under the null hypothesis:

p-value: P(t_{n-2}\leq -3.82) + P(t_{n-2}\geq  3.82)

As you can see to calculate it you need the information of the sample size to determine the degrees of freedom of the distribution.

If you want to use the rejection region approach, the sample size is also needed to determine the critical values.

But since this test is two tailed at α: 0.05 and there was a confidence interval with confidence level 0.95 (which is complementary to the level of significance) you can use it to decide whether to reject the null hypothesis.

Using the CI, the decision rule is as follows:

If the CI includes the "zero", do not reject the null hypothesis.

If the CI doesn't include the "zero", reject the null hypothesis.

The calculated interval for the slope is: [-0.11; -0.04]

As you can see, both limits of the interval are negative and do not include the zero, so the decision is to reject the null hypothesis.

At a 5% significance level, you can conclude that the relationship between the year and the number of earthquakes above magnitude 7.0 is statistically significant.

I hope this helps!

(full output in attachment)

5 0
4 years ago
In Ms. Chance's first period class, there are 9 students with hazel eyes, 10 with brown eyes, 7 with blue eyes, and 2 students w
MA_775_DIABLO [31]
Brown eyes because most students have brown eyes. It would be less likely to get a student with blue or green eyes.
5 0
3 years ago
Directions: Calculate the area of a circle using 3.14x the radius
Leokris [45]

\qquad\qquad\huge\underline{{\sf Answer}}♨

As we know ~

Area of the circle is :

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

<h3>Problem 1</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 4.4\div 2

\qquad \sf  \dashrightarrow \:r = 2.2 \: mm

Now find the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.2)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {4.84}^{}

\qquad \sf  \dashrightarrow \:area  \approx 15.2 \:  \: mm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 2</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 3.7 \div 2

\qquad \sf  \dashrightarrow \:r = 1.85 \:  \: cm

Bow, calculate the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (1.85) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 3.4225 {}^{}

\qquad \sf  \dashrightarrow \:area  \approx 10.75 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 3 </h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (8.3) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 68.89

\qquad \sf  \dashrightarrow \:area \approx216.31 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 4</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 5.8 \div 2

\qquad \sf  \dashrightarrow \:r = 2.9 \:  \: yd

now, let's calculate area ~

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.9)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  8.41

\qquad \sf  \dashrightarrow \:area  \approx26.41 \:  \: yd {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 5</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 1 \div 2

\qquad \sf  \dashrightarrow \:r = 0.5 \:  \: yd

Now, let's calculate area ~

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (0.5) {}^{2}

\qquad \sf  \dashrightarrow \:3.14  \times 0.25

\qquad \sf  \dashrightarrow \:area \approx0.785 \:  \: yd {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 6</h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(8)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 64

\qquad \sf  \dashrightarrow \:area = 200.96 \:  \: yd {}^{2}

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

8 0
2 years ago
Can someone please help me with this, im doing a test and im stuck on this question.
Mashutka [201]

Answer:

20

Step-by-step explanation:

first do 75/50 = 1.5

then multiply red by 1.5

13 landed on red so

13 x 1.5 = 19.5

but since you round up it is 20

6 0
3 years ago
What is the expanded form 845,212
navik [9.2K]
845.212= 8*100.000+4*10.000+5*1.000+2*100+1*10+2
5 0
3 years ago
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