What is the area of a triangle with vertices at (−4, 1) , (−7, 5) , and (0, 1) ?

Mallory needs to go to the airport. It takes her 120 minutes to get there by car. This

sergeinik [125]

Answer:

120(3)=x

120*3=360

x=360 min to go to the Airport by train.

Step-by-step explanation:

6 0

3 years ago

Abbot is paid weekly for every hour he works. He worked 6 hours each day for 5 days last week. He earned $408, before taxes. Wri

Contact [7]

Answer:

p=13.6h

Step-by-step explanation:

if he works 6 hours a day for five days that's 30 hours. $408 divided by 30 hours is $13.6 per hour.

3 0

3 years ago

Suppose you pay back $575 on a $525 loan you had for 75 days. What was your simple annual interest rate? State your result to th

KengaRu [80]

The simple annual interest rate for the $ 525 loan is equal to 46.35 %.

<h3>What is the interest rate behind a pay back?</h3>

In this situation we assume that the loan does not accumulate interests continuously in time. Hence, the interest rate for paying the loan back 75 days later is:

575 = 525 · (1 + r/100)

50 = 525 · r /100

5000 = 525 · r

r = 9.524

The loan has an interest rate of 9.524 % for 75 days. Simple annual interest rate is determine by rule of three:

r' = 9.524 × 365/75

r' = 46.350

The simple annual interest rate for the $ 525 loan is equal to 46.35 %.

To learn more on interests: brainly.com/question/26457073

#SPJ1

7 0

2 years ago

You have 6 dollars +12 pennies +to quarters + 4 dimes=2 nickels how much do you have

alina1380 [7]

You may have $26 I not a genius but this is easy

3 0

3 years ago

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

IgorLugansk [536]

Answer:

(a) 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

Step-by-step explanation:

We are given that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.75.

(a) Also, the average porosity for 20 specimens from the seam was 4.85.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.85

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 20

$\mu$ = true average porosity

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.85-1.96 \times {\frac{0.75}{\sqrt{20} } }$ , $4.85+1.96 \times {\frac{0.75}{\sqrt{20} } }$ ]

= [4.52 , 5.18]

Therefore, 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) Now, there is another seam based on 16 specimens with a sample average porosity of 4.56.

The pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.56

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 16

$\mu$ = true average porosity

Here for constructing 98% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 98% confidence interval for the true mean, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98 {As the critical value of z at 1% level

of significance are -2.3263 & 2.3263}

P(-2.3263 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.3263$ ) = 0.98

P( $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.56-2.3263 \times {\frac{0.75}{\sqrt{16} } }$ , $4.56+2.3263 \times {\frac{0.75}{\sqrt{16} } }$ ]

= [4.12 , 4.99]

Therefore, 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

7 0

3 years ago

What is the area of a triangle with vertices at (−4, 1) , ​ (−7, 5) ​ , and ​ (0, 1) ​?

What is the area of a triangle with vertices at (−4, 1) , (−7, 5) , and (0, 1) ?