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3 years ago

Help needed here. So Please help.

Mathematics

2 answers:

matrenka [14]3 years ago

7 0

It’s the compensation

earnstyle [38]3 years ago

5 0

Answer:

I notice in the compensation strategy it's much simpler than basic subtraction.

Step-by-step explanation:

Just say that ;)

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If you spin each of the following spinners once, what is the probability of getting red and an even number?

vichka [17]

The colored one would be a 1 in 4 and the b&w is a 1 in 5 :1

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3 years ago

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What is the final step to be performed when computing σ(x - 2)2??

jasenka [17]

<span>σ</span> is a symbol for summation that means that it is used to add all the vales of the equation in the range for example taking 0<x<4 and x is an integer so Ď(x - 2)^2= (1-2)^2+(2-2)^2+(3-2)^2 The options were a. Square each value b. Subtract 2 points from each score C. Sum the squared values d. Subtract 22 from each X2 value As we can see above the answer is C. Sum the squared values

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3 years ago

Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean

earnstyle [38]

Answer:

$t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167$

$p_v =2*P(t_{23}$

We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.

Step-by-step explanation:

Data given

$n_1 =10$ represent the sample size for group 1

$n_2 =15$ represent the sample size for group 2

$\bar X_1 =1085$ represent the sample mean for the group 1

$\bar X_2 =1034$ represent the sample mean for the group 2

$s_1=52$ represent the sample standard deviation for group 1

$s_2=61$ represent the sample standard deviation for group 2

We are assuming that the two samples are normally distributed with equal variances and that means:

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

System of hypothesis

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

The statistic is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The degrees of freedom are given by:

$n_1+n_2 -2$

And the pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

Replacing we got:

$S^2_p =\frac{(10-1)(52)^2 +(15 -1)(61)^2}{10 +15 -2}=3323.043$

And the deviation would be:

$S_p=57.646$

The degrees of freedom are:

$df=10+15-2=23$

The statistic would be:

$t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167$

The p value would be

$p_v =2*P(t_{23}$

We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.

6 0

3 years ago

HELP! I'LL GIVE BRAINLIEST!!!

Nikolay [14]

4-c
5-a
6-c
7- im not sure

7 0

4 years ago

Solve them please, correct answers will be marked as brainliest & a follow :) thank you x<br>

DaniilM [7]

Answer:

this is the answer that I could do please check it if I could help you

4 0

3 years ago