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2 years ago

8

Statistics show that the sales force of Golden Wholesalers successfully closed 1,711 sales out of 1,950 sales calls. What was th

eir percent success rate?

1 answer:

Arisa [49]2 years ago

8 0

1711 sales out of 1950 calls. The success rate is quite impressive: 1711/1950=87.7%

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All I need is this answer,please help!!

il63 [147K]

Hello there,

There is a total of 40 animals in that in 10 zebras, 10 lions, 10 elephants, and 10 monkeys.
So the probability of them both being lions is pretty rare. 2/40 so that would give you a 1/20 chance of both crackers being both lions.

I hope this helps!

7 0

2 years ago

The following data shows the low temperature in Vancouver, in degrees Celsius, for two weeks in January one year. 8.90, 8.35, 8.

frutty [35]

There is no option of the box plots, so I have created a version that would represent this data.

To make the box plot you will need the lower extreme, lower quartile, median, upper quartile, and upper extreme.

Please see the attached picture.

5 0

2 years ago

A package of hamburgers contains 8 patties and costs $7.50. Part A.) Luna has to buy at least 16 packages for an upcoming picnic

Rus_ich [418]

Answer:

Part A) $p\geq 128\ patties$

Part B) She will spend more than $\$142.50$

Step-by-step explanation:

Let

p-----> the number of hamburger patties

Part A) Luna has to buy at least 16 packages for an upcoming picnic

$p\geq 16*8$

$p\geq 128\ patties$

Part B) Suppose she actually needs more than 150 hamburgers. How much will she spend?

Let

c---------> the total cost

step 1

Divide 150 hamburgers by 8 (a package of hamburgers)

so

$\frac{150}{8} =18.75\ package$

round to the nearest whole number

$18.75=19\ package$ ----> the minimum number of packages

step 2

$c>19*\$7.50$

$c>\$142.50$

7 0

2 years ago

A quality control technician works in a factory that produces computer monitors. Each day, she randomly selects monitors and tes

jeka94

Answer:

There is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 300

p = 5% = 0.05

Alpha, α = 0.05

Number of dead pixels , x = 24

First, we design the null and the alternate hypothesis

$H_{0}: p = 0.05\\H_A: p > 0.05$

This is a one-tailed(right) test.

Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{24}{300} = 0.08$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.08-0.05}{\sqrt{\frac{0.05(1-0.05)}{300}}} = 2.384$

Now, we calculate the p-value from excel.

P-value = 0.00856

Since the p-value is smaller than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.

5 0

1 year ago

Kent has two similar cylindrical pipes, Pipe A and Pipe B. The radius of Pipe A is 6 cm, and the radius of Pipe B is 2 cm. What

Alchen [17]

Answer: C)27:1

Step-by-step explanation:

Given, Kent has two similar cylindrical pipes, Pipe A and Pipe B. The radius of Pipe A is 6 cm, and the radius of Pipe B is 2 cm.

Volume of cylinder = $\pi r^2h$ , where r= radius and h = height.

Also, If two figures are similar then ratio of volume is equal to the cube of any dimension .

The ratio of the volume of Pipe A to the volume of Pipe B is given by :-

$\dfrac{\text{Volume of pipe A}}{\text{Volume of pipe B}}=\dfrac{6^3}{2^3}\\\\=\dfrac{216}{8}=\dfrac{27}{1}$

Thus, the ratio of the volume of Pipe A to the volume of Pipe B = 27:1

So, the correct option is C)27:1.

7 0

2 years ago