All categories

3 years ago

If h(x) = 5 x and k(x)=1/x, which expression is equivalent to (koh)(x)?

2 answers:

4 0

May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x

1) If h(x) = 5x and k(x) = 1/x

Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)

2) If h(x) = 5 + x and k (x) = 1/x

Then (k o h)(x) =k ( h(x) ) = k (5+x) = 1 / [5 + x]

stealth61 [152]3 years ago

4 0

Answer:

The answer is B!! 1/(5+x)

Step-by-step explanation:

You might be interested in

An appliance manufacturer stockpiles washers and dryers in a large warehouse for shipment to retail stores. Sometimes in handlin

Over [174]

Answer:

This is strong evidence that more than 2% get damaged

Step-by-step explanation:

The company goal is to keep the proportion of damaged machines below 2%

$H_o: p = 0.02$

$H_a: p > 0.02$

One day an inspector randomly checks 66 washers and finds that 6 of them have scratches or dents

So, x = 6

n = 66

$\widehat{p}=\frac{x}{n}$

$\widehat{p}=\frac{6}{66}$

Formula : $z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z=\frac{\frac{6}{66}-0.02}{\sqrt{\frac{0.02(1-0.02)}{66}}}$

$z=4.114$

So, the value of test statistic is 4.114

2) What is the P-value of the test statistic?

So, p value using calculator = .00002

p value < α

So, we accept the alternate hypothesis

So, This is strong evidence that more than 2% get damaged

8 0

3 years ago

Consider the quadratic function, f(x)=-2x^2+12x+32=0. Select all statements that are true for the function. The value of the dis

Andrews [41]

Answer:

Step-by-step explanation:

-2x² + 12x + 32 = 0

discriminant = 12² - 4(-2)(32) = 400

positive discriminant means there are two distinct real roots

x = [-12 ± √(12² – 4(-2)·32)] / [2·(-2)]

= [-12 ± √400] / (-4)

= [-12 ± 20] /(-4)

= 8, -2

8 0

3 years ago

Read 2 more answers

Let the universal set, s, have 66 elements. a and b are subsets of s. set a contains 19 elements and set b contains 34 elements.

Otrada [13]

The cardinality of the union between two sets is given by

$N(A\cup B) = N(A) + N(B) - N(A \cap B)$

In fact, if an element is in both A and B, it is counted in both N(A) and N(B), so we're counting it twice. By subtracting the cardinality of the interserction, we're taking back one of these "extra-counted" element, and the count is correct. In your case, the numbers are

$N(A \cup B) = 19+34-10 = 43$