Answer:
Since the null hypothesis is true, finding the significance is a type I error.
The probability of the year I error = level of significance = 0.05.
so, the number of tests that will be incorrectly found significant is computed as follow: 0.05 * 100 = 5
Therefore, 5 tests will be incorrectly found significant given that the null hypothesis is true.
Base=x+7 Height=x Area=60. 60=((x+7)(x))/2----->60=(x^2+7x)/2
120=x^2+7x X^2+7X-120=0. QUDRATIC FORMULA: -7+/- SQRT(49+480)/2
(-7+/-23)/2... 8 and -15. -15 ISNT the answer because you cant have a negative length. X=8 Base=8+7---->15 Height=8
Answer:
Hope this helps
Step-by-step explanation:
a) <e and <c
b) <c and <b
c) <c and <a
d) <c and <b, <c and <d, <d and <e, <e and <a, <a and <b
e) c = 30, a = 90, b= 60, e= 30, d= 150
Answer:
1st answer
Step-by-step explanation:
First you subrtract 3 from 0 then add 4 to -3
Answer:
59 to 66
Step-by-step explanation:
Mean test scores = u = 74.2
Standard Deviation =
= 9.6
According to the given data, following is the range of grades:
Grade A: 85% to 100%
Grade B: 55% to 85%
Grade C: 19% to 55%
Grade D: 6% to 19%
Grade F: 0% to 6%
So, the grade D will be given to the students from 6% to 19% scores. We can convert these percentages to numerical limits using the z scores. First we need to to identify the corresponding z scores of these limits.
6% to 19% in decimal form would be 0.06 to 0.19. Corresponding z score for 0.06 is -1.56 and that for 0.19 is -0.88 (From the z table)
The formula for z score is:

For z = -1.56, we get:

For z = -0.88, we get:

Therefore, a numerical limits for a D grade would be from 59 to 66 (rounded to nearest whole numbers)