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AlexFokin [52]
3 years ago
15

A​ chef's specialty calls for ​one- fiftieth of an ounce of a spice per serving. The chef has a digital scale that shows the wei

ght as a decimal. What scale reading does the chef need to see for this spice when preparing the specialty for 43 ​people?
Mathematics
1 answer:
Anna35 [415]3 years ago
3 0

Answer:

The scale reading that the chef need to see for this spice when preparing the specialty for 43 people is 0.86.

Step-by-step explanation:

In order to find the answer, you need to multiply the amount that the chef uses per serving for the number of servings. The statement indicates that the chef uses ​one- fiftieth of an ounce of a spice per serving and this is represented as 1/50 and you have to multiply this for 43 that is the number of people.

You can find 1/50 as a decimal dividing 1 by 50:

1/50=0.02

Now, you can multiply this value for 43:

0.02=43=0.86

According to this, the answer is that the scale reading that the chef need to see for this spice when preparing the specialty for 43 people is 0.86.

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Given ASTU and ADEF, what is mZU?
salantis [7]

Answer:

Maybe the same as the angle of F (46)

Step-by-step explanation:

If you see the two shapes are the same. but different size. so therfore the angles are same.  

3 0
3 years ago
Find the range of the data set 247, 366, 785, 997
nasty-shy [4]
Range is the difference between the lowest and highest values. 
997- 247= 750
8 0
3 years ago
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Snowcat [4.5K]

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-1

Step-by-step explanation:

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5 0
3 years ago
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(1+(0.065÷365))^365t=4
marta [7]
This is the future value quadrupled in t years at an annual interest rate of 6.5% compounded daily.  We need to find t.

1*(1+0.065/365)^(365t)t=4
take log on both sides,
365t(log(1+0.065/365)=log(4)
=>
365t=log(4)/log(1+0.065/365)
t=(log(4)/log(1+0.065/365))/365
=(1.38629/.000178066)/365
=21.33 years

Check with the rule of 69, applicable to continuous compounding (an approximation to current problem) to double money, it take 69/interest rate in % years.
=69/6.5
=10.62 years
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7 0
3 years ago
At a particular chess club, it is quite common for a chess game to take over two hours to complete. Suppose that the lengths of
Serga [27]

Answer:

A game would need to be at least 257.67 minutes long to qualify for the Endurance Board.

257.67 minutes = 257 minutes, 40 seconds = 4 hours, 17 minutes, 40 seconds.

Step-by-step explanation:

Games that are given special recognition on the Endurance Board are the games that last in the longest 1% of all games.

If X is the random variable that represents the time a chess game takes before it is completed.

X is said to be normally distributed with

Mean = μ = 153 minutes

Standard deviation = σ = 45 minutes

Let games that last the longest 1% of the time last for a minimum of x' minutes.

P(X > x') = 1% = 0.01

P(X ≤ x') = 1 - P(X > x') = 1 - 0.01

P(X ≤ x') = 0.99

Indicating that such games are longer than 99% of all chess games.

This is a normal distribution problem

Let the z-score for these type of longest games with a minimum duration of x' minutes be z'.

P(X ≤ x') = P(z ≤ z') = 0.99

From the normal distribution table, z' = 2.326

z-score of any value is given as the value minus the mean divided by the standard deviation.

z = (x - μ)/σ

So,

z' = (x' - μ)/σ

2.326 = (x' - 153)/45

x' = (2.326×45) + 153

x' = 104.67 + 153 = 257.67 minutes = 257 minutes, 40 seconds.

Hope this Helps!!!

7 0
3 years ago
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