Amarketing research analyst has collected data from a random sample of 100 customers who have purchased a particular “coupon special.” the 100 people spent an average of $24.57 in the store with a standard deviation of $6.60. before performing this sampling the marketing manager claimed that average purchase by those responding to the coupon offer would be at least $25.00. what is the test t-value to two places and can the manager’s hypothesis (the null hypothesis) be rejected at the 5% significance level?

he kicked it up? is this the answer ?

step-by-step explanation:

explanation:

infrared radiation has wavelength in the range between 700 nm and 1 mm. lets' convert these values into scientific notation, keeping in mind that:

let's start from the lower value, we have:

(we moved the dot two places to the left, so we increased the exponent by 2 units, from -9 to -7)

the higher value is:

so, the range of wavelength of infrared radiation is:

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb