Write an equation for the line that is perpendicular to the line y= -9 that passes through the point (-3,4).

Jenna is a profesional typist. She can type 22,500 words in 5 hours of typing. In one hour she can type _____ words. In one minu

Gennadij [26K]

She can type 4500 in 1 hour and 75 in a minute .

5 0

3 years ago

Need help with this

svlad2 [7]

Answer:

What is questions?

Step-by-step explanation:

What is questions?

6 0

2 years ago

The amount of time, in minutes, that a woman must wait for a cab is uniformly distributed between zero and 12 minutes, inclusive

murzikaleks [220]

Answer:

$P(X\leq x) =\frac{x-a}{b-a}, a \leq x \leq b$

And using this formula we have this:

$P(X$

Then we can conclude that the probability that that a person waits fewer than 11 minutes is approximately 0.917

Step-by-step explanation:

Let X the random variable of interest that a woman must wait for a cab"the amount of time in minutes " and we know that the distribution for this random variable is given by:

$X \sim Unif (a=0, b =12)$

And we want to find the following probability:

$P(X$

And for this case we can use the cumulative distribution function given by:

$P(X\leq x) =\frac{x-a}{b-a}, a \leq x \leq b$

And using this formula we have this:

$P(X$

Then we can conclude that the probability that that a person waits fewer than 11 minutes is approximately 0.917

7 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

zavuch27 [327]

Answer:

Step-by-step explanation:

Hey B is the correct answer as the x is negative and y is positive

Hope this helps!!!

3 0

3 years ago

Read 2 more answers

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago