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vladimir2022 [97]
2 years ago
15

Evaluate the expression for the given value of x. 6x + 7 for x = 2 The solution is

Mathematics
1 answer:
pav-90 [236]2 years ago
4 0

Answer:

6x + 7

6 * 2

12 + 7

=19

Step-by-step explanation:

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2 years ago
Tensile-strength tests were carried out on two different grades of wire rod (Fluidized Bed Patenting of Wire Rods, Wire J., June
Shtirlitz [24]

Answer:

t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569

p_v =P(t_{256}>28.569) \approx 0

So with the p value obtained and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.

Step-by-step explanation:

The statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2 +10

Alternative hypothesis: \mu_1 >\mu_2 +10

Or equivalently:

Null hypothesis: \mu_1 - \mu_2 \leq 10

Alternative hypothesis: \mu_1 -\mu_2>10

Our notation on this case :

n_1 =129 represent the sample size for group AISI 1078

n_2 =129 represent the sample size for group AISI 1064

\bar X_1 =123.6 represent the sample mean for the group AISI 1078

\bar X_2 =107.6 represent the sample mean for the group AISI 1064

s_1=2.0 represent the sample standard deviation for group 1 AISI 1078

s_2=1.3 represent the sample standard deviation for group AISI 1064

And now we can calculate the statistic:

t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569

Now we can calculate the degrees of freedom given by:

df=129+129-2=256

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

p_v =P(t_{256}>28.569) \approx 0

So with the p value obtained and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.  

7 0
3 years ago
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