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4 years ago

A cyclist travels 10 km from home to the park.

Mathematics

1 answer:

lesantik [10]4 years ago

4 0

Answer:

20 km/hr

Step-by-step explanation:

The total time she takes to travel 10 km is 30 minutes of 0.5 hours.

To calculate speed, we can divide the distance traveled by the time it takes.

$s = \frac{10}{0.5} = 20$ km/hr

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How many pounds of cashews that cost $14 per pound must be mixed with 5 pounds of peanuts that cost $6.50 per pound to make mixe

Fantom [35]

The amount of cashew mixed is 5 pounds

<h3>Word problem on total cost</h3>

Cost of 1 pound of cashew = $14

Cost of 1 pound of peanut = $6.50

Total cost of 1 pound of the mixed nuts = $10.25

Cost of 5 pounds of peanuts = 5 x $6.50

Cost of 5 pounds of peanuts = $32.50

Let the amount of cashew be x

Total cost of x cashew = 14x

Total amount of the mixed nuts = x + 5

Total cost of the mixed nuts = 10.25(x+5)

Total cost of the mixed nuts = 10.25x + 51.25

14x + 32.50 = 10.25x + 51.25

14x - 10.25x = 51.25 - 32.50

3.75x = 18.75

x = 18.75/3.75

x = 5

Therefore, the amount of cashew mixed is 5 pounds

Learn more on word problem on cost here: brainly.com/question/21405634

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8 0

2 years ago

Read 2 more answers

Susan went shopping to purchase new clothes. She

arlik [135]

Answer:

13.375

Step-by-step explanation:

you have to add 28 ± 17 = 45 ± 8.857 = 13.375

6 0

3 years ago

How to attach image <br><br> sure sure

Bogdan [553]

Answer:

Press the kinda paper clip thing it says attach file then press that

Step-by-step explanation:

3 0

2 years ago

A cookie factory monitored the number of broken cookies per pack yesterday.

trapecia [35]

Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

sqrt(37.69) = 6.139

s*sqrt(n-1) = 18.2

$s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } }$ $\leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }$

confidence interval:

(18.2/7.948) < S < (18.2/6.139)

2.290 < S < 2.965

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3 years ago

Adding and subtracting real numbers - find each difference - 1.7 - (-3.8)

shepuryov [24]

Answer:

2.1

Step-by-step explanation:

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3 years ago