Whats the difference of 12-33/4

This is the expanded form (7 x 10) (4 X 1) +(8 X ) +( 9 X ) WHAT IS THAT IN STANDARD FORM AND WORD FORM

alex41 [277]

70(4)+...

8 times what and 9 times what

7 0

3 years ago

Evaluate if x =3 and y = -2: x squared + 2 y A. 5 B. 10 C. 2 D. 3

Luden [163]

Of you have 3 squared you get 9 then 2 times -2 is -4. then you add them 9=-4 which would be ...A: 5

5 0

4 years ago

Read 2 more answers

Math Help Please?

lapo4ka [179]

You share only one distribution, so we'll focus on that one: mean: 30; std. dev.: 4.

Draw a "standard normal curve." Draw a vertical line in the exact middle of your curve. Label this line "30." Now "one standard dev. above the mean" is 30+4=34; "two std. devs. above the mean is 30+4+4=38, or 30+8=38. "three std. devs. above the mean is 30+3(4) = 42.

Now work in the other direction. Start with the mean: 30. But now subtract the std. dev. (4) instead of adding it. You'll get 30-4=26. This is "1 std. dev. below the mean. Continue: find 2 and 3 std. devs. below the mean.

8 0

3 years ago

HELP PLEASEEEEEEEEEEEEEEEEEEEEEEEE

Karo-lina-s [1.5K]

<h2>Answer:</h2>

18

Step-by-step explanation:

10$ 180$

10 x 18 = 180

so she work 18 hours.

7 0

3 years ago

A new post-surgical treatment is being compared with a standard treatment. Seven subjects receive the new treatment, while seven

AnnyKZ [126]

Answer:

$t=\frac{17.71-28.71}{\sqrt{\frac{4.461^2}{7}+\frac{7.387^2}{7}}}}=-3.372$

$df=n_{S}+n_{N}-2=7+7-2=12$

Since is a one sided test the p value would be:

$p_v =P(t_{(12)}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean time for the new treatment is significantly lower than the time for the standard treatment at 5% of significance.

Step-by-step explanation:

New: 12, 13,15,19,20,21

Standard: 18,23,24,30,32,35

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Data given and notation

$\bar X_{N}=17.71$ represent the mean for the new case

$\bar X_{S}=28.71$ represent the mean for the standard case

$s_{N}=4.461$ represent the sample standard deviation for the new case

$s_{S}=7.387$ represent the sample standard deviation for the standard case

$n_{S}=7$ sample size selected for the standard case

$n_{N}=7$ sample size selected for the new case

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if recovery time for patients who receive the new treatment is less than the mean for patients who receive the standard treatment, the system of hypothesis would be:

Null hypothesis: $\mu_{N} \geq \mu_{S}$

Alternative hypothesis: $\mu_{N} < \mu_{S}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{N}-\bar X_{S}}{\sqrt{\frac{s^2_{N}}{n_{N}}+\frac{s^2_{S}}{n_{S}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{17.71-28.71}{\sqrt{\frac{4.461^2}{7}+\frac{7.387^2}{7}}}}=-3.372$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{S}+n_{N}-2=7+7-2=12$

Since is a one sided test the p value would be:

$p_v =P(t_{(12)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean time for the new treatment is significantly lower than the time for the standard treatment at 5% of significance.

6 0

4 years ago