Find the slope of a line through (-1,1) and (0,4)

What is 16 to 29 and 15 in at ratio

Semmy [17]

16:29:15 I think that would be it but, I'm really sorry if that's not right

7 0

2 years ago

PLEASEEEE HELPPP

soldier1979 [14.2K]

Ik I’m not know help but gee I forgot this just go on google

4 0

2 years ago

Read 2 more answers

Diego has 90 song on his playlist If 40% of them are hip-hop how many songs Are hip-hop

abruzzese [7]

Answer:

225

Step-by-step explanation:

i used a calculater and did 90 divided by .40

8 0

3 years ago

Write the null and alternative hypotheses in words and then symbols for each of the following situations.

Marrrta [24]

Answer:

A) the hypothesis in words and symbols are;

Null hypothesis: Grades have not improved for most students

Alternative hypothesis: Grades have improved for most students

Null hypothesis; H0: p^ = 0.5

Alternative hypothesis; Ha: p^ > 0.5

B) Null hypothesis: During march madness, employees averaged 15 minutes per day on non - business activities

Alternative hypothesis: During march madness, employees averaged more than 15 minutes per day on non-business activities

Null hypothesis; H0: μ = 15

Alternative hypothesis; Ha: μ > 15

Step-by-step explanation:

A) We are told that they sampled 200 of the students who used their service in the past year and ask them if their grades have improved or declined from the previous year. Thus, there will be a 50% chance of success for their grades to have improved and a 50% chance for it to remain the same.

Thus; p^ = 0.5

So the hypothesis in words and symbols are;

Null hypothesis: Grades have not improved for most students

Alternative hypothesis: Grades have improved for most students

Null hypothesis; H0: p = 0.5

Alternative hypothesis; Ha: p > 0.5

B) We are told that they estimate that on a regular business day employees spend on average 15 minutes of company time checking personal email, making personal phone calls, etc

Also, They also collect data on how much company time employees spend on such non-business activities during March Madness.

The hypotheses in words and symbols are;

Null hypothesis: During march madness, employees averaged 15 minutes per day on non - business activities

Alternative hypothesis: During march madness, employees averaged more than 15 minutes per day on non-business activities

Null hypothesis; H0: μ = 15

Alternative hypothesis; Ha: μ > 15

6 0

3 years ago

A researcher wishes to estimate the average blood alcohol concentration (BAC) for drivers involved in fatal accidents who are f

Mnenie [13.5K]

Answer:

A 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

Step-by-step explanation:

We are given that a researcher randomly selects records from 60 such drivers in 2009 and determines the sample mean BAC to be 0.16 g/dL with a standard deviation of 0.080 g/dL.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean BAC = 0.16 g/dL

s = sample standard deviation = 0.080 g/dL

n = sample of drivers = 60

$\mu$ = population mean BAC in fatal crashes

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

So, a 90% confidence interval for the population mean, $\mu$ is;

P(-1.672 < $t_5_9$ < 1.672) = 0.90 {As the critical value of t at 59 degrees of

freedom are -1.672 & 1.672 with P = 5%} P(-1.672 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.672) = 0.90

P( $-1.672 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $0.16-1.672 \times {\frac{0.08}{\sqrt{60} } }$ , $0.16+1.672 \times {\frac{0.08}{\sqrt{60} } }$ ]

= [0.143, 0.177]

Therefore, a 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

7 0

3 years ago