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AnnZ [28]
2 years ago
8

If you have 60 million barrels and you have 14% kerosene and 34% gasoline what is the total production of gasoline and kerosene

combined barrels
Mathematics
2 answers:
Drupady [299]2 years ago
6 0
48% or 48/100 only in percents and fractions
In-s [12.5K]2 years ago
4 0

Answer:

Total production of gasoline and kerosene combined 48% that is 28.8 million.

Step-by-step explanation:

Total production = 60 million barrels

Gasoline = 34%

Kerosene = 14%

Total combined = 34% + 14% = 48%

48% of 60 million = 0.48 × 60 = 28.8 million

Total production of gasoline and kerosene combined 48% that is 28.8 million.

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trasher [3.6K]
Ummm... £1/£8 I think
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2 years ago
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Emilce pide 5 platos por semana para almorzar en la oficina. El menú ofrece 5 opciones diferentes. De cuantas maneras puede real
MrRissso [65]

Answer:

120 different ways

Step-by-step explanation:

For the first lunch, Emilce has 5 options to choose,

In the second lunch, now she has only 4 options, as one was already chosen.

Then, in the third lunch, she has 3 options.

In the fourth lunch, she has 2 options.

In the last lunch, she has just 1 option.

So, the number of different ways she can order her lunches in the week is:

5 * 4 * 3 * 2 * 1 = 120

She can order in 120 different ways.

5 0
2 years ago
0.58 divided into 0.5568
Fiesta28 [93]
The answer to that would be 1.04166667.
3 0
2 years ago
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of
Thepotemich [5.8K]
<h2><u>Answer with explanation</u>:</h2>

As per given , we have

sample size : n= 65

degree of freedom : df=n-1=64

sample mean : \overline{x}=19.5

sample standard deviation : s= 5.2

Since , the population standard deviation is not given  , so we apply t-test.

Significance level  for 90% confidence : \alpha=1-0.90=0.10

t-critical value for significance level 0.10 and df = 64 would be :

t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.6690

Formula for Confidence interval :

\overline{x}\pm t_c\dfrac{s}{\sqrt{n}}

Then , 90%  confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

19.5\pm(1.669)\dfrac{5.2}{\sqrt{65}}

=19.5\pm1.076

(19.5-1.076,\ 19.5+1.076)=(18.424,\ 20.576)

∴ 90% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.424, 20.576)

Significance level  for 95% confidence : \alpha=1-0.95=0.05

t-critical value for significance level 0.05 and df = 64 would be :

t_c=t_{\alpha/2, df}=t_{0.05,\ 64}=1.9977

Then , 95%  confidence intervals for the population mean number of weekly customer contacts for the sales personnel would be :

19.5\pm(1.9977)\dfrac{5.2}{\sqrt{65}}

=19.5\pm1.288

(19.5-1.288,\ 19.5+1.288)=(18.212,\ 20.788)

∴ 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel : (18.212, 20.788)

4 0
2 years ago
Andrea is shopping for items for her pet sitting business because the store is having a sale where everything in the store is 5%
mars1129 [50]

Answer:

i promise it's A i am 1,00000000000000000000000000000000000000 sure

Step-by-step explanation:

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2 years ago
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