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ikadub [295]
3 years ago
15

A police officer believes red cars speed more often than any other color car. During his 25 shifts this month, he records the nu

mber of speeding tickets he writes to red and non-red cars.
a. Determine whether the following statement is true or false.

1. The police officer is conducting an experiment.

b. All of the following variables could be confounding variables EXCEPT for which one?

a. time of day
b. color of car
c. location
d. the police officer
Mathematics
1 answer:
AysviL [449]3 years ago
3 0

Answer:

Step-by-step explanation:

a. Yes, the following statement is true. The police officer is gathering data by writing down all the speeding tickets and seperating them into two categories, red and non-red cars. He can later analyze this data and compare them to determine whether or not his hypothesis is true.

b. The color of the car is the only variable that would not be a confounding variable in this scenario. This would instead be the independent variable and does not in any way affect the dependent variable which would be the number of speeding tickets received.

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Step-by-step explanation:

You would first find the area of the whole rectangle without the hole, so (15)(30)=450

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To find the area you must subtract the area of the hole from the rectangle.

450-80=370

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3 years ago
Evaluate the following pic
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Answer:

1) \sqrt{1225}+\sqrt{1024}=67

2)  \sqrt[3]{-1331}=-11

3) Evaluating 2:p :: p:8 we get p=\pm 4

4) x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}

5) \frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6

Step-by-step explanation:

1) \sqrt{1225}+\sqrt{1024}

Prime factors of 1225 : 5x5x7x7

Prime factors of 1024: 2x2x2x2x2x2x2x2x2x2

\sqrt{1225}+\sqrt{1024}\\=\sqrt{5\times5\times7\times7}+\sqrt{2\times2\times2\times2\times2\times2\times2\times2\times2\times2}\\=\sqrt{5^2\times7^2}+\sqrt{2^2\times2^2\times2^2\times2^2\times2^2}\\=5\times7+(2\times2\times2\times2\times2)\\=35+32\\=67

\sqrt{1225}+\sqrt{1024}=67

2) \sqrt[3]{-1331}

We know that \sqrt[n]{-x}=-\sqrt[n]{x} \ ( \ if \ n \ is \ odd)

Applying radical rule:

\sqrt[3]{-1331}\\=-\sqrt[3]{1331} \\=-\sqrt[3]{11\times\11\times11}\\=-\sqrt[3]{11^3} \\Using \ \sqrt[n]{x^n}=x \\=-11

So, \sqrt[3]{-1331}=-11

3) 2:p :: p:8

It can be written as:

p*p=2*8\\p^2=16\\Taking \ square \ root \ on \ both \ sides\\\sqrt{p^2}=\sqrt{16}\\p=\pm 4

Evaluating 2:p :: p:8 we get p=\pm 4

4) x^3+y^2+z \ when \ x=3, y=-2, x=-6

Put value of x, y and z in equation and solve:

x^3+y^2+z \\=(3)^3+(-2)^2+(-6)\\=27+4-6\\=25

So, x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}

5) \frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}

We know (-a)^n = (a)^n when n is even and (-a)^n = (-a)^n when n is odd

\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}\\\\=\frac{1296\times-8\times 27}{46656}\\\\=\frac{-279936}{46656} \\\\=-6

So, \frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6

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