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2 years ago

carmen hiked at Yosemite National Park for 2.75 hours. Her average speed was 3.1 miles per hour. How many miles did she hike

Mathematics

1 answer:

SVEN [57.7K]2 years ago

5 0

Answer: 8.525 miles

Step-by-step explanation:

I got this by multiplying 3.1 by 2.75

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Product day is an 8 oz bottle of cough medicine that sells for $1.36 product b is a 16 oz bottle of cough medicine that cost $3.

Kazeer [188]

Answer: 8 oz bottle

Step-by-step explanation:

$1.36 / 8 oz = $0.17 / oz

$3.20 / 16 oz = $0.20 / oz

$0.17 < $0.20

5 0

2 years ago

Over the first five years of owning her car, Gina drove about 12,700 miles the first year, 15,478 miles the second year, 12,675

kvv77 [185]

Answer:

Part a)

Mean: 13155.6

Mode:None

Median: 12700

Step-by-step explanation:

12700 : 1

15478: 2

12675: 3

11850: 4

13075: 5

Mean :

$=\frac{Sum of all values}{Total no. of values} \\=\frac{12700 + 15478 + 12675 + 11850 + 13075}{5}\\=\frac{65778}{5}\\= 13155.6$

Mode:

All values are distinct, therefore there is no mode. A mode is value which occurs most frequently.

Median:

Arrange the values in ascending order:

11850 , 12675, 12700, 12075, 15478

The mid value is median: 12700

Part b)

It depends upon the data. In this question the best way to predict how many miles Gina will drive in the next years is by using Median of the data.

The reason is that there is a potential outlier in the data i.e 15478. In case of outlier median tends to give better predictions.

6 0

3 years ago

the sum of three numbers is 131. The second of three numbers is seven more than twice the first. The third number is 12 less tha

spin [16.1K]

No. 1 = x
No. 2 = 2x+7
No. 3 = x-12

Answer:

x+(2x+7)+(x-12) = 131

From there you solve to get the numbers:

34, 75 and 22

3 0

2 years ago

Suppose you are given either a fair dice or an unfair dice (6-sided). You have no basis for considering either dice more likely

hoa [83]

Answer: Our required probability is 0.83.

Step-by-step explanation:

Since we have given that

Number of dices = 2

Number of fair dice = 1

Probability of getting a fair dice P(E₁) = $\dfrac{1}{2}$

Number of unfair dice = 1

Probability of getting a unfair dice P(E₂) = $\dfrac{1}{2}$

Probability of getting a 3 for the fair dice P(A|E₁)= $\dfrac{1}{6}$

Probability of getting a 3 for the unfair dice P(A|E₂) = $\dfrac{1}{3}$

So, we need to find the probability that the die he rolled is fair given that the outcome is 3.

So, we will use "Bayes theorem":

$P(E_1|A)=\dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\\\\(E_1|A)=\dfrac{0.5\times 0.16}{0.5\times 0.16+0.5\times 0.34}\\\\P(E_1|A)=0.83$

Hence, our required probability is 0.83.

8 0

3 years ago

Hiiii can u please pls pls pls

Shkiper50 [21]

Answer:

x | y

0 | 0

6 | 2

12 | 4

Step-by-step explanation:

Multiply each x value in the table by 1/3 to get 0, 2, and 4 for your y values.

4 0

2 years ago

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