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3 years ago

7

Which situation describes a car that can travel $40 miles per gallon of gasoline?

1 answer:

Salsk061 [2.6K]3 years ago

8 0

Answer:

A car can travel 80 miles on 2 gallons of gasoline.

Step-by-step explanation:

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Please helppppppppppppp

aleksley [76]

Answer:

The circumference is 18.84 inches.

Step-by-step explanation:

To find the circumference of the circle we can use the formula:

$C=2\pi r$

Where C is circumference and r is radius.

Now that we have the formula, we can substitute the numbers:

$C=2(3.14*3)\\C=2(9.42)\\C=18.84$

8 0

2 years ago

Read 2 more answers

Select the correct answer.

Liono4ka [1.6K]

Answer:

Option C.

Step-by-step explanation:

It is given that,

Total number of applicants = 9

Number of job = 3

We need to select 3 applicants out of 9. So, the required number of ways is

$^9P_3=\dfrac{9!}{(9-3)!}$

$^9P_3=\dfrac{9\times 8\times 7\times 6!}{6!}$

$^9P_3=9\times 8\times 7$

$^9P_3=504$

Therefore, there are $^9P_3=504$ ways the positions can be filled because the order in which the applicants are chosen matters.

Hence, the correct option is C.

4 0

2 years ago

NEED HELP PLEASEEEE!

vladimir1956 [14]

Answer:

181.76

Step-by-step explanation:

The calculator gave me this answer

5 0

3 years ago

The 7 in 72 represents how many times the value of the 7 in 78?

Grace [21]

The answer is zero times.

3 0

2 years ago

(Anderson, 1.14) Assume that P(A) = 0.4 and P(B) = 0.7. Making no further assumptions on A and B, show that P(A ∩ B) satisfies 0

Goshia [24]

Answer with Step-by-step explanation:

We are given that

P(A)=0.4 and P(B)=0.7

We know that

$P(A)+P(B)+P(A\cap B)=P(A\cup B)$

We know that

Maximum value of $P(A\cup B)$ =1 and minimum value of $P(A\cup B)$ =0

$0\leq P(A\cup B )\leq 1$

$0\leq P(A)+P(B)-P(A\cap B)\leq 1$

$0\leq 0.4+0.7-P(A\cap B)\leq 1$

$0\leq 1.1-P(A\cap B)\leq 1$

$0\leq 1.1-P(A\cap B)$

$P(A\cap B)\leq 1.1$

It is not possible that $P(A\cap B)$ is equal to 1.1

$1.1-P(A\cap B)\leq 1$

$-P(A\cap B)\leq 1-1.1=-0.1$

Multiply by (-1) on both sides

$P(A\cap B)\geq 0.1$

Again, $P(A\cup B)\geq P(B)$

$0.4+0.7-P(A\cap B)\geq 0.7$

$1.1-P(A\cap B)\geq 0.7$

$-P(A\cap B)\geq -1.1+0.7=-0.4$

Multiply by (-1) on both sides

$P(A\cap B)\leq 0.4$

Hence, $0.1\leq P(A\cap B)\leq 0.4$

3 0

3 years ago