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2 years ago

Paul can type 60 words per minute and Jennifer can type 80 words per minute. How does Paul’s typing speed compare to Jennifer’s

Mathematics

2 answers:

Harrizon [31]2 years ago

5 0

Jennifer can type 1/3 faster than Paul
or
Paul can type at 75% the speed of Jennifer

yan [13]2 years ago

4 0

I'm not certain how exactly this question is supposed to be answered.

If it wants a ratio, Paul and Jennifer's speeds would be in a ratio of 3:4.

It also could be Paul types 20 words per minute slower than Jennifer.

It could be Paul types 33% slower than Jennifer.

Another answer could be Paul's typing speed is 3/4ths of Jennifer's typing speed.

Feel free to ask any further questions.

You might be interested in

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

2 years ago

A digital rain gauge has an outdoor sensor that collects rainfall and transmits data to an indoor display. Assume you produced a

evablogger [386]

Answer:

a. A continuous graph

b. a straight line graph with a gentle positive slope

c. The graph consists of three straight lines attached end to end, with the first segment being an inclined positively sloped graph, the second segment is an horizontal line while the third segment is a gentle positively sloped straight line

Step-by-step explanation:

a. A continuous graph

Given that the rain gauge is a digital rain gauge, the values transmitted to the indoor display are those collected by the outdoor sensor, we have that the output to the transmitter can be any value within a possible and finite range of values, therefore the graph is continuous

b. Whereby it rained at a constant rate of 0.1 inch per hour over 24-hour period, we have;

i) Whereby the graph is the level of water in the rain gauge over a given period of time, the shape of the graph will be a straight line graph with a gentle positive slope of 0.1 the domain will be from 0 to 24 hours, while the range will be from 0 to 2.4 inches

c) i) Given that i) the rain falls at a constant rate of 0.1 inch per hour for 6 hours, and ii) over the next 6 hours it stops raining, after which iii) it rained 0.2 inch for the next 12 hours, we have;

i) In the first part The graph will be a straight line graph with a gentle positive slope of 0.1, up to the point (6, 0.6)

ii) When the rain stops for 6 hours, from the point (6, 0.6) to the point (12, 0.12) the graph will be an horizontal line

iii) From the point (12, 0.12) to the point (24, 0.24), is represented on the graph by a straight line with a gentle positive slope of 0.2.

7 0

3 years ago

The average speed of greyhound dogs is about 18.4 meters per second. A particular greyhound breeder claims that her dogs are fas

irinina [24]

Answer:

The calculated value Z = 1.183 < 1.96 at 0.05 level of significance

Null hypothesis is accepted

A particular greyhound breeder claims that her dogs are faster than the average greyhound

Step-by-step explanation:

Step(i):-

Given the average speed of greyhound dogs is about 18.4 meters per second.

Size of the sample 'n' = 35

mean of the sample x⁻ = 18.7

Population standard deviation = 1.5m/s

level of significance (∝) = 0.05

Step(ii):-

Null hypothesis : H₀ : μ = 18.4

Alternative hypothesis H₁ : μ ≠ 18.4

Test statistic

$Z = \frac{x^{-} - mean}{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{18.7-18.4}{\frac{1.5}{\sqrt{35} } }$

Z = 1.183

Conclusion:-

The calculated value Z = 1.183 < 1.96 at 0.05 level of significance

Null hypothesis is accepted

A particular greyhound breeder claims that her dogs are faster than the average greyhound

8 0

2 years ago

Solve: −8x−7/ 3 =11

VladimirAG [237]

Answer: $x=-\frac{5}{3}$

Step-by-step explanation:

Given the equation $-8x-\frac{7}{3}=11$ , you need find the value of the variable x.

Then, you have to solve for x:

Add $\frac{7}{3}$ to both sides of the equation.

Divide by -8 to both sides of the equation.

Once you do this proccedure, you will get the value of the variable x.

Then:

$-8x-\frac{7}{3}+\frac{7}{3}=11+\frac{7}{3}$

$-8x=\frac{40}{3}\\\\\frac{-8x}{-8}=\frac{\frac{40}{3}}{-8}\\\\x=-\frac{5}{3}$

8 0

3 years ago

1. The local furniture store pays $140 for a chair and sells it with a

zalisa [80]

Answer:

a. $56

b. $196

Step-by-step explanation:

a. Markup price

= 40% of 140

= 0.40*140

= $56

b. Selling Price

= $140 + $56

= $196

5 0

3 years ago