Suppose the time that it takes a certain large bank to approve a home loan is normally distributed with mean (in days) μ and standard deviation σ=1. the bank advertises that it approve loans in 5 days, on average, but measurements on a random sample of 500 loan applications to this bank gave a mean approval time of = 5.3 days. is this evidence that the mean time to approval is actually longer than advertised? to answer this, test the hypotheses h0: μ=5, ha: μ> 5 at significance level α=0.01. you conclude that: a. ha should be rejected. b. there is a 5% chance that the null hypothesis is true. c. h0 should be rejected. d. h0 should not be rejected.

(c) H0 should be rejected

Step-by-step explanation:

Null hypothesis (H0): population mean is equal to 5

Alternate hypothesis (Ha): population mean is greater than 5

Z = (sample mean - population mean) ÷ (sd/√n)

sample mean = 5.3, population mean = 5, sd = 1, n = 500

Z = (5.3 - 5) ÷ (1/√500) = 0.3 ÷ 0.045 = 6.67

Using the normal distribution table, for a one tailed test at 0.01 significance level, the critical value is 2.326

Conclusion:

Since 6.67 is greater than 2.326, reject the null hypothesis (H0)

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1

dilation scale factor 2: starting with δabc, draw the dilation image of the triangle with a center at the origin and a scale factor of two. notice that every coordinate of the original triangle has been multiplied by the scale factor (x2). dilation with scale factor 2, multiply by 2.