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3 years ago

Please help me with 7-12

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

3 0

Answer:

-5

Step-by-step explanation:

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Multiply. Enter the product in singlets form. 10/18x9/10=

Alexeev081 [22]

Answer:

1/2

Step-by-step explanation:

6 0

3 years ago

The table represents the total miles traveled, y, after a number of hours, x. Hours, x Miles, y 2.5 150 4.0 240 5.5 330 7.0 420

maw [93]

(2.5,150)(4,240)
slope = (240 - 150) / (4 - 2.5) = 90 / 1.5 = 60

y = mx + b
slope(m) = 60
(4,240)...x = 4 and y = 240
now we sub and find b, the y int
240 = 60(4) + b
240 = 240 + b
240 - 240 = b
0 = b

so ur equation is : y = 60x + 0 which is written as : y = 60x <==

7 0

3 years ago

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Problem 1.1<br> Solve this equation:<br> 2(x + 5) = 3x + 1

Tamiku [17]

Answer:

x = 9

Step-by-step explanation:

Distribute

2x+2*5=3x+1

2x+10=3x+1

Subtract 3x from both sides

-x+10=1

Subtract 10 from both sides

-x=-9

Multiply both sides by -1

x=9

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3 years ago

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AOB and BOC are _____.

Elanso [62]

The answer is supplementary.

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3 years ago

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Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. The

Tems11 [23]

Answer:

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , 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 $1 - \frac{\alpha}{2}$ .

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that $n = 387, \pi = \frac{298}{387} = 0.77$

84% confidence level

So $\alpha = 0.16$ , z is the value of Z that has a p-value of $1 - \frac{0.16}{2} = 0.92$ , so $Z = 1.405$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8$

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

6 0

3 years ago