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Pepsi [2]
3 years ago
13

I need help on number 28

Mathematics
2 answers:
Gala2k [10]3 years ago
6 0
-4 & -3 is your answer because -4+-3=-7 and -4 times -3 = 12
8_murik_8 [283]3 years ago
5 0
The two integers are -3, and -4. -3 added to -4 equals -7. -3 times -4 equals 12 as a negative times a negative is a positive.



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Find the value of f(2)
sergiy2304 [10]

Answer:

We have, y=f(x)

From graph, when x = 2, y =f(x)=-5. So, functional value at x=2, f(2) = -5

3 0
3 years ago
Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yi
aalyn [17]

Answer:

a) t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222  

df=12+15-2=25  

p_v =P(t_{25}>6.222) =8.26x10^{-7}

So with the p value obtained and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

b) (91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309

(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691

Step-by-step explanation:

Notation and hypothesis

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2  

And the statistic is given by this formula:  

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

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:  

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

This last one is an unbiased estimator of the common variance \sigma^2  

Part a

The system of hypothesis on this case are:  

Null hypothesis: \mu_2 \leq \mu_1  

Alternative hypothesis: \mu_2 > \mu_1  

Or equivalently:  

Null hypothesis: \mu_2 - \mu_1 \leq 0  

Alternative hypothesis: \mu_2 -\mu_1 > 0  

Our notation on this case :  

n_1 =12 represent the sample size for group 1  

n_2 =15 represent the sample size for group 2  

\bar X_1 =85 represent the sample mean for the group 1  

\bar X_2 =91 represent the sample mean for the group 2  

s_1=3 represent the sample standard deviation for group 1  

s_2=2 represent the sample standard deviation for group 2  

First we can begin finding the pooled variance:  

\S^2_p =\frac{(12-1)(3)^2 +(15 -1)(2)^2}{12 +15 -2}=6.2  

And the deviation would be just the square root of the variance:  

S_p=2.490  

Calculate the statistic

And now we can calculate the statistic:  

t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222  

Now we can calculate the degrees of freedom given by:  

df=12+15-2=25  

Calculate the p value

And now we can calculate the p value using the altenative hypothesis:  

p_v =P(t_{25}>6.222) =8.26x10^{-7}

Conclusion

So with the p value obtained and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

Part b

For this case the confidence interval is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

For the 99% of confidence we have \alpha=1-0.99 = 0.01 and \alpha/2 =0.005 and the critical value with 25 degrees of freedom on the t distribution is t_{\alpha/2}= 2.79

And replacing we got:

(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309

(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691

7 0
3 years ago
4(1/2+3/4)+(-2)<br><br> I need help
nirvana33 [79]

Answer:

3

Step-by-step explanation:

4(1/2+3/4)+(-2)   Distribute.

2+3+(-2)             Simplify.

2+3-2

5-2

3

Hope this helps!! Have an amazing day (。・∀・)ノ゙

5 0
2 years ago
Round 13.045 to the nearest whole number.
Daniel [21]

Answer:

13. it says nearest whole number.

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
The force in Newtons acting on a particle located x meters from the origin is given by the function F(x)=12x2+7. What is the wor
Shkiper50 [21]

W = \displaystyle \int^{5}_{2}   F(x) ~ dx\\\\\\= \displaystyle \int^{5}_{2} \left(12x^2 +7\right) ~dx\\\\\\= 12\displaystyle \int^{5}_{2} x^2 ~dx + \displaystyle \int^{5}_{2} 7~ dx\\\\\\=12 \left[\dfrac{x^3}3 \right]^{5}_{2}  + 7\left[x\right]^{5}_{2}\\\\\\=4(5^3-2^3) + 7(5-2) = 468 + 21 = 489~ J

5 0
2 years ago
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