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3 years ago

What is the slope?? what is the y-intercept

Mathematics

2 answers:

kumpel [21]3 years ago

5 0

Y-intercept is 1

slope is 2

USPshnik [31]3 years ago

3 0

M= 2
y=mx+b
1=2(1)+b
1=2+b
-2 -2
-1=b
so this equation is y=2x-1

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oceanside bike rental shop charges a $17 fixed fee plus six dollars an hour for renting a bike. Tim paid $47 to rent a bike. How

gavmur [86]

Approximately 8 hours

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2 years ago

The accompanying data represent the daily (for example, Monday to Tuesday) movement of Johnson & Johnson (JNJ) stock for

egoroff_w [7]

Supposing that the stock increases in 37 days, the 95% confidence interval for the proportion of days JMJ stock increases is: (0.484, 0.7292)

The lower bound is of 0.484.
The upper bound is of 0.7292.

The interpretation is that we are 95% sure that the true proportion of all days in which the JMJ stock increases is between 0.484 and 0.7292.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

Supposing that it increases on 37 out of 61 days:

$n = 61, \pi = \frac{37}{61} = 0.6066$

95% confidence level

So $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 - 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.484$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 + 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.7292$

The 95% confidence interval for the proportion of days JMJ stock increases is (0.484, 0.7292), in which 0.484 is the lower bound and 0.7292 is the upper bound.

The interpretation is that we are 95% sure that the true proportion of all days in which the JMJ stock increases is between 0.484 and 0.7292.

A similar problem is given at brainly.com/question/16807970

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2 years ago

Mark buys a used car for $12,000. The value of the car depreciates 10% per year from the time he bought the car.

timama [110]

Answer:

option D

Step-by-step explanation:

r = 10% = 0.10

$v(t) = P (1 - r)^t\\\\V(t) = 12000(1 - 0.10)^t\\\\V(t) = 12000\times (0.9)^t$