Heights were measured for a random sample of 10 plants grown while being treated with a particular nutrient. The sample mean and sample standard deviation of those height measurements were 46 centimeters and 7 centimeters, respectively. Assume that the population of heights of treated plants is normally distributed with mean μ. Based on the sample, can it be concluded that μ is different from 45 centimeters? Use the 0.05 level of significance.

The calculated value   t = 0.452< 2.2621 at 0.05 level of significance

null hypothesis is accepted

Based on the sample, can it be concluded that μ is not different from 45 centimeters

Step-by-step explanation:

Step( i ):-

Heights were measured for a random sample of 10 plants

Size of the sample 'n' = 10

Mean of the sample (x⁻ ) = 46 centimeters

Standard deviation of the sample (s) = 7 centimeters

Mean of the Population ( μ ) = 45

Step(ii):-

Null Hypothesis :H₀:( μ ) = 45

Alternative Hypothesis : H₁:  μ ) ≠ 45

Test statistic

           t = 0.452

  Degrees of freedom

                 γ = n-1 = 10 -1 = 9

t₀.₀₅ =  2.2621

The calculated value   t = 0.452< 2.2621 at 0.05 level of significance

null hypothesis is accepted

Based on the sample, can it be concluded that μ is not different from 45 centimeters

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3