Need help how do I Solve a=2x+6xz for x.

Two chemists working for a chicken fast-food company, have been producing a very popular sauce. Let’s call then Jesse and Mr. Wh

a_sh-v [17]

Answer:

Check the explanation

Step-by-step explanation:

Let $\overline{x}$ and $\overline{y}$ be sample means of white and Jesse denotes are two random variables.

Given that both samples are having normally distributed.

Assume $\overline{x}$ having with mean $\mu_{1}$ and $\overline{y}$ having mean $\mu_{2}$

Also we have given the variance is constant

A)

We can test hypothesis as

$H0: \mu_{1} = \mu_{2}H1: \mu_{1} > \mu_{2}$

For this problem

Test statistic is

$T=\frac{\overline {x}-\overline {y}}{s\sqrt{\frac {1}{n1} +\frac{1}{n2}}}$

Where

$s=\sqrt{\frac{(n1-1)*s1^{2}+(n2-1)*s2^{2}}{n1+n2-2}}$

We have given all information for samples

By calculations we get

s=2.41

T=2.52

Here test statistic is having t-distribution with df=(10+7-2)=15

So p-value is P(t15>2.52)=0.012

Here significance level is 0.05

Since p-value is <0.05 we are rejecting null hypothesis at 95% confidence.

We can conclude that White has significant higher mean than Jesse. This claim we can made at 95% confidence.

3 0

3 years ago

Will the border line of the graph of y < - x2 + 4x + 5 be a solid line or a broken

Free_Kalibri [48]

Step-by-step explanation:

When the inequality uses < or >, it is a broken line.

For the point (0, 0) it means that x = 0 and y = 0.

Since (0) < (0)² + 4(0) + 5 is true, the point is part of the solution.

7 0

2 years ago

Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the p

PIT_PIT [208]

Answer:

Colin has 8 sheets left for his third class.

Step-by-step explanation:

Given that:

Total Number of pieces of papers = $x$

Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad

Writing the equation:

$\text{Number of pieces of papers used for 1st class =} \dfrac{x}{2} -5 ...... (1)$

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

$\text{Number of pieces of papers used for 2nd class =} \dfrac{x}{2} -5+2 = \dfrac{x}2 -3 ...... (2)$

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

$\text{number of pieces of papers left for the third class = }x-(\dfrac{x}{2}-5)-(\dfrac{x}{2}-3)\\\Rightarrow x-\dfrac{x}2-\dfrac{x}2+5+3\\\Rightarrow x-x+5+3\\\Rightarrow 8$