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2 years ago

7: A flight lasting 3 hours and 45 minutes lands at 7:50 pm. What time did the flight take off?

Mathematics

2 answers:

Alexandra [31]2 years ago

7 0

Around 4 pm because 7 - 3 = 4.

luda_lava [24]2 years ago

7 0

4:05 pm

Step-by-step explanation:

take off time 7:50 - 3:45

= 4:05

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PLEASE HELP! Which interval indicates the answers most students gave?

xenn [34]

30-39 is the interval because it has the most responses

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3 years ago

Student scores on exams given by a certain instructor have mean 74 and standard deviation 14. This instructor is about to give t

Lapatulllka [165]

Answer:

$P(\bar X >80)=P(Z>2.143)=1-P(z$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable that represent the Student scores on exams given by a certain instructor, we know that X have the following distribution:

$X \sim N(\mu=74, \sigma=14)$

The sampling distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

The deduction is explained below we have this:

$E(\bar X)= E(\sum_{i=1}^{n}\frac{x_i}{n})= \sum_{i=1}^n \frac{E(x_i)}{n}= \frac{n\mu}{n}=\mu$

$Var(\bar X)=Var(\sum_{i=1}^{n}\frac{x_i}{n})= \frac{1}{n^2}\sum_{i=1}^n Var(x_i)$

Since the variance for each individual observation is $Var(x_i)=\sigma^2$ then:

$Var(\bar X)=\frac{n \sigma^2}{n^2}=\frac{\sigma}{n}$

And then for this special case:

$\bar X \sim N(74,\frac{14}{\sqrt{25}}=2.8)$

We are interested on this probability:

$P(\bar X >80)$

And we have already found the probability distribution for the sample mean on part a. So on this case we can use the z score formula given by:

$z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

Applying this we have the following result:

$P(\bar X >80)=P(Z>\frac{80-74}{\frac{14}{\sqrt{25}}})=P(Z>2.143)$

And using the normal standard distribution, Excel or a calculator we find this:

$P(Z>2.143)=1-P(z$

6 0

4 years ago

You guys I need help with something like this again

Over [174]

In general an annual rate of $r$ compounded monthly will multiply the amount by a factor of $1+\frac{r}{12}$ each month.

Each month of 4.3% is a factor of $1+\frac{.043}{12}$ . That goes for five months, so is multiplied by itself five times, so gets a fifth power.

Each month of 13.7% is a factor of $1+\frac{.137}{12}$ . He pays this for 12-5=7 months, so it gets an exponent of seven.

We started the year owing $2600 of principal, so after the year we owe

$2600(1+\frac{.043}{12})^5(1+\frac{.137}{12})^7$ .

Choice A.

8 0

3 years ago

Please help !!!!! idk how to do this

vladimir1956 [14]

Answer:

Step-by-step explanation:

When x equals -3, then the first equation is simplified to 9 + 6 + 3 = 18; and the second equation confirms the same answer by showing -6 x -3 = 18

7 0

3 years ago

The average of the weights of six people sitting in a boar is 145 pounds. After a seventh person gets into the boat, the average

Tems11 [23]

Weight : 

⇒ x for 6 people
⇒ y for the 7th person

$x/6=145 \\ (x+y)/6=147 \\ \\ x=145*6 \\ x=870 \\ \\ x+y=1029 \\ \\ 870+y=1029 \\ y=1029-870 \\ y=159$

The correct answer is B).

5 0

3 years ago

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