We know that a sample of 1800 computer chips revealed that 53% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 54% of the chips do not fail in the first 1000 hours of their use.

We get that:

$n=1800\\\\p_1=53\%=0.53\\\\p_2=54\%=0.54\\$

We calculate the standar deviation:

$\sigma=\sqrt{\frac{0.53 \cdot (1-0.53)}{n}}=\sqrt{\frac{0.53 \cdot 0.47}{1800}}=0.01176$

We calculate the value of the test statistic:

$z=\frac{p_1-p_2}{\sigma}=\frac{0.53-0.54}{0.01176}=-0.85$

So the value of the test statistic is -0.85.

5 0

4 years ago

Matt is showing his work in simplifying (6.2 − 1.6) − 4.4 + 7.8. Identify any error in his work or reasoning.

shusha [124]

Start with the parentheses first

4.6 -4.4+7.8 then calculate from left to right
= 0.2+7.8 = 8

6 0

3 years ago

A printer needs to make a poster that will have a total of 500 cm2 that will have 3 cm margins on the sides and 2 cm margins on

goblinko [34]

Answer:

The largest printed area is 381.43 cm^2 .

Step-by-step explanation:

The width of the full poster = 500/x

Length of the printed area = l =x - 4

Width of the printed area = w = (500/x) - 2

Area of the printed space = (x - 4) × ((500/x) - 2)

Now, take derivative of the area

A' = (x - 4)×(-500/x^2) + ((500/x) - 2)

A' = (2000/x^2) - 2

A' = (2000-2x^2) / x^2

put derivative equal to zero like A' = 0

(2000-2x^2) / x^2 = 0

2000 = 2x^2

x^2 = 1000

x = 31.62

So, the length of the original poster is = x = 31.62 cm

The width of the full poster = 500/x = 15.81 cm

l = x - 4 = 27.62

w = 500/x - 2 = 13.81

Therefore, The area of space available for printing is = l × w

= 27.62 × 13.81 = 381.43 cm^2 .

5 0

3 years ago

Hal has two cans containing biscuit dough. The cans are cylindrical and their dimensions (in inches) are as follows.

poizon [28]

C: The volume of can B is four times the volume of can A

8 0

3 years ago