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murzikaleks [220]
3 years ago
7

What is 50g/cm= kg/m

Mathematics
2 answers:
Anni [7]3 years ago
6 0
The answer is 5 kg/m
yulyashka [42]3 years ago
5 0
<span>The correct answer is 5 kg / m</span>
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Rafael writes an expression to represent “a number, n, decreased by thirty-one.” Then he evaluates that expression if n=29. Whic
yulyashka [42]
This is a subtraction expression.o evaluate, substitute 29 for the variable in the expression.to<span>evaluate the expression, subtract 31 from 29.</span>
3 0
3 years ago
Read 2 more answers
Iron Man Training While training for an iron man competition, Tony bikes for 60 miles and runs for 15 miles. If his bikes speed
ivolga24 [154]

Answer:

<u>Tony biked 1 hour 40 minutes and run 3 hours 20 minutes.</u>

Step-by-step explanation:

How long did Tony speed on his bike?

Let s to be Tony's running speed, then 8s is Tony's biking speed

Let's recall that Time = Distance/Speed

Bike time + Running time = 5 hrs

Replacing with the real values, we have:

60/8s + 15/s = 5

8s as Lowest Common Denominator

60 + 8 * 15 = 5 * 8s

60 + 120 = 40s

-40s = -180

s = -180-40

s = 4.5 ⇒ 8s = 36

Tony run at 4.5 mph and biked at 36 mph

Now, we can calculate the time, this way:

Bike time = Bike distance/Bike speed

Replacing with the real values, we have:

Bike time = 60/36

Bike time = 5/3 (Dividing by 12)

5/3 hours * 60 = 100 minutes or 1 hour 40 minutes

Running time = 5 hours - 1 hour 40 minutes = 3 hours 20 minutes

<u>Tony biked 1 hour 40 minutes and run 3 hours 20 minutes.</u>

4 0
3 years ago
Help me plssssssssss
Dmitry_Shevchenko [17]

Answer:

just do the thing and then thing the do

Step-by-step explanation:

4 0
3 years ago
Write the word sentence as a equation. Then solve the equation 13 subtracted from a number w is 15
patriot [66]
W-13=15 add 13 on both sides to isolate variable and you get w=13+15, w=28
7 0
2 years ago
Suppose management could improve the process by reducing the mean time required for an oil change (but keeping the standard devi
Alja [10]

This question is incomplete, the complete question is;

The owners of Spiffy Lube want to offer their customers a 10-minute guarantee on their standard oil change service. If the oil change takes longer than 10 minutes to complete, the customers is given a coupon for a free oil change at the next visit. Based on past history, the owners believe that the timer required to complete an oil change has a normal distribution with a mean of 8.6 minutes and a standard deviation of 1.2 minutes.

Suppose management could improve the process by reducing the mean time required for an oil change (but keeping the standard deviation the same). How much change in the mean service time would be required to allow for a 10-minute guarantee that gives a coupon to no more than 1 out of every 25 customers on average

Answer:

Required change in the mean service time is 7.8988

Step-by-step explanation:

Given the data in the question;

How much change in the mean service time would be required to allow for a 10-minute guarantee that gives a coupon to no more than 1 out of every 25 customers on average

let mean = μ

p( x > 10 ) ≤ (1/25)

p( x > 10 ) ≤ 0.4

p( x-μ / 1.2  > 10-μ / 1.2 ) ≤ 0.4

(10-μ / 1.2 ) ≤ 0.4

(10-μ / 1.2 ) ≥ q_{norm} ( 0.96 )

(10-μ / 1.2 ≥ 1.751

10-μ  = ≥ 1.751 × 1.2

10-μ  ≤ 2.1012

μ ≤ 10 - 2.1012

μ ≤ 7.8988

Therefore, required change in the mean service time is 7.8988

8 0
3 years ago
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