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3 years ago

Which one is it A. B. C. or D.

Mathematics

2 answers:

zvonat [6]3 years ago

8 0

The answer would be C

11Alexandr11 [23.1K]3 years ago

3 0

The answer is C. It says that c(weight) = the cost

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In onenote show how many positive integers less than 50 are multiples of 4, but not multiples of 6

dimaraw [331]

Multiples of 4 = 4 8 12 16 20 24 28 32 36 40 44 48

After Removing multiples of 6= 4 8 16 20 28 32 40 44

5 0

3 years ago

How do you solve this?

Novay_Z [31]

Let's solve your equation step-by-step.

−7x=−2x2+15

Step 1: Subtract -2x^2+15 from both sides.

−7x−(−2x2+15)=−2x2+15−(−2x2+15)

2x2−7x−15=0

Step 2: Factor left side of equation.

(2x+3)(x−5)=0

Step 3: Set factors equal to 0.

2x+3=0 or x−5=0

x= $\frac{-3}{2}$ or x=5

Answer:

x= $\frac{-3}{2}$ or x=5

Hope this helps

7 0

3 years ago

By driving 8 mph faster than Bob, John can make a 230 mile trip in one half hour less. How fast does Bob drive on the trip? Roun

____ [38]

Answer: The speed of Bob is 56.8 km/hr.

Step-by-step explanation:

Let the speed of Bob be 'x'.

Let the speed of John be 'x-8'.

Distance covered = 230 miles

time = $1\dfrac{1}{2}=\dfrac{3}{2}\ hr$

According to question, we get that

$\dfrac{230}{x}-\dfrac{230}{x+8}=\dfrac{3}{2}\\\\230\dfrac{x+8-x}{x(x+8)}=\dfrac{3}{2}\\\\\dfrac{230\times 8}{x^2+8x}=\dfrac{3}{2}\\\\\dfrac{1840}{x^2+8x}=\dfrac{3}{2}\\\\1840\times 2=3x^2+24x\\\\3680=3x^2+24x\\\\3x^2+24x-3680=0\\\\x\approx 56.8\ km/hr$

Hence, the speed of Bob is 56.8 km/hr.

6 0

3 years ago

10 points What is the slope of the function y 4x - 6 Type your answer

ki77a [65]

Answer:

Step-by-step explanation:

6 0

3 years ago

Let F be the event that a student is enrolled in a finance course, and let S be the event that a student is enrolled in a statis

Inessa [10]

Answer:

0.375

Step-by-step explanation:

Given that F is the event that a student is enrolled in a finance course, and let S be the event that a student is enrolled in a statistics course.

It is known that 40% of all students are enrolled in a finance course and 35% of all students are enrolled in statistics. Included in these numbers are 15% who are enrolled in both statistics and finance.

P(F) = 0.40: P(S) = 0.35: P(FS) = 0.15

A student is randomly selected, and it is found that the student is enrolled in finance.

Probability that this student is also enrolled in statistics/ student is enrolled in finance.

= $\frac{P(FS)}{P(F)} \\=\frac{0.15}{0.40} \\=0.375$

(Using conditional probability formula)

5 0

3 years ago