All categories

2 years ago

Plz help me with number 8 plz help me I’m begging you plz help me I need your help plz help me I’m begging you plz help me

Mathematics

2 answers:

valina [46]2 years ago

8 0

seal dive, s is greater than 990ft and less than 1850 ft

=》 990<s<1850

vampirchik [111]2 years ago

6 0

Answer:

B. 990<s<1,850

You might be interested in

Can someone please solve this

Svetradugi [14.3K]

Answer:

The answer is the 3rd one.

Step-by-step explanation:

3 times 3 is 9

4 times 4 is 16

13 is in between those 2 numbers

4 0

2 years ago

Read 2 more answers

Solve each equation below and write the answer on the line. Plan to spend 6 minutes total on this question. What is X?

zepelin [54]

A) x= -15/7
B) x= -2
C) x= -3
D) x= 3

6 0

2 years ago

Through:(4,2)parallel to x=0

bogdanovich [222]

A line that is parallel to x=0 has to be x=k where k is a constant. In (4,2), x=4. Therefore, the answer is x=4.

4 0

2 years ago

A simple random sample of 28 Lego sets is obtained and the number of pieces in each set was counted.The sample has a standard de

Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

Given that:

A simple random sample n = 28

sample standard deviation S = 12.65

standard deviation $\sigma$ = 11.53

Level of significance ∝ = 0.05

The objective is to test the claim that the number of pieces in a set has a standard deviation different from 11.53.

The null hypothesis and the alternative hypothesis can be computed as follows:

Null hypothesis:

$H_0: \sigma^2 = \sigma_0^2$

Alternative hypothesis:

$H_1: \sigma^2 \neq \sigma_0^2$

The test statistics can be determined by using the following formula in order to test if the claim is statistically significant or not.

$X_0^2 = \dfrac{(n-1)S^2}{\sigma_0^2}$

$X_0^2 = \dfrac{(28-1)(12.65)^2}{(11.53)^2}$

$X_0^2 = \dfrac{(27)(160.0225)}{132.9409}$

$X_0^2 = \dfrac{4320.6075}{132.9409}$

$X_0^2 = 32.5002125$

$X^2_{1- \alpha/2 , df} = X^2_{1- 0.05/2 , n-1}$

$X^2_{1- \alpha/2 , df} = X^2_{1- 0.025 , 28-1}$

From the chi-square probabilities table at 0.975 and degree of freedom 27;

$X^2_{0.975 , 27}$ = 14.573

$X^2_{\alpha/2 , df} = X^2_{ 0.05/2 , n-1}$

$X^2_{\alpha/2 , df} = X^2_{0.025 , 28-1}$

From the chi-square probabilities table at 0.975 and degree of freedom 27;

$X^2_{0.025 , 27}=$ 43.195

Decision Rule: To reject the null hypothesis if $X^2_0 \ > \ X^2_{\alpha/2 , df} \ \ \ or \ \ \ X^2_0 \ < \ X^2_{1- \alpha/2 , df}$ ; otherwise , do not reject the null hypothesis:

The rejection region is $X^2_0 \ > 43.195 \ \ \ or \ \ \ X^2_0 \ < \ 14.573$

Conclusion:

We fail to reject the null hypothesis since test statistic value 32.5002125 lies between 14.573 and 43.195.

6 0

2 years ago

Order theses from least to greatest

agasfer [191]

5.75, 5.751, 59/10, 5 10/11 pls mark brainliest if it helped

5 0

1 year ago