All categories

3 years ago

Which number is closet to 51.4 is it 51.41,51.39 ,51.041 ,51.402

Mathematics

2 answers:

lyudmila [28]3 years ago

8 0

The number closest to 51.4 is 51.402

irga5000 [103]3 years ago

8 0

Answer:

The correct option is D) 51.402

Step-by-step explanation:

Consider the provided number 51.4

We need to identify the number closet to 51.4

The provided options are: 51.41, 51.39, 51.041, 51.402

To find which is closet number simply find the difference between the provided number and options given, As the difference will decrease the number gets closer.

Find the difference between 51.41 and 51.4

51.41-51.4=0.01

Find the difference between 51.4 and 51.39

51.4-51.39=0.01

Find the difference between 51.4 and 51.041

51.4-51.041=0.359

Find the difference between 51.402 and 51.4

51.402-51.4=0.002

Here, 0.002 is the minimum difference that means the number 51.402 is the closet to 51.4

Hence, the correct option is D) 51.402

You might be interested in

I really really need help with this please.

11111nata11111 [884]

The number of ancestors going back through the 5th generation, including Tle-nle and counting Tle-nle as the 1st generation is:

= 1 + 3 + 3^2 + 3^3 + 3^4
= (3^5 - 1) / (3 - 1)
= 242 / 2
= 121

Since we included Tle-nle as the 1st generation, we will only compute up to the 4th power. If it is until the 6th generation, add 3^5 to the equation.

8 0

3 years ago

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)