In an article appearing in today’s health a writer states that the average number of calories in a serving of popcorn is 75. to determine if the average number of calories in a serving of popcorn is different from 75, a nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean number of calories per serving to be 78 with a sample standard deviation of 7. at the a = .05 level of significance, does the nutritionist have enough evidence to reject the writer’s claim?

The nutritionist have enough evidence to reject the writer’s claim .

Step-by-step explanation:

Claim : The average number of calories in a serving of popcorn is different from 75

Null hypothesis :

Alternate hypothesis :  

n = 20

Sample standard deviation s = 7

Since n < 30 and population standard deviation is not given .

So, we will use t - test .

Formula :

Substitute the values

So, t calculated = 1.9166

Degree of freedom = n-1 = 20-1 = 19

Refer the t table

So,

Since t critical > t calculated

So, we are fail to reject the null hypothesis .

So,  the nutritionist do not have enough evidence to reject the writer’s claim .

im sorry but no. i just need to answer for 50 poonts oof

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1