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LUCKY_DIMON [66]
2 years ago
6

If you need to use 4/5 of a cup of strawberries for a single pie . How many cups would be needed for 2 pies.

Mathematics
1 answer:
kirza4 [7]2 years ago
5 0

Answer:

1 3/5      or 8/5     cups for 2 pies

Step-by-step explanation:

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How do you write 53,806 in expanded form using exponents
densk [106]
(5x10^4)+(3x10^3)+(8x10^2)+(0x10^1)+6x10^0)
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3 years ago
Air-USA has a policy of booking as many as 24 persons on an airplane that can seat only 22. (Past studies have revealed that onl
Alex17521 [72]

Answer:

B. no, it is not low enough

A. no, it is not low enough

Step-by-step explanation:

Given that Air-USA has a policy of booking as many as 24 persons on an airplane that can seat only 22.

Prob for  a random person booked arrive for flight = 0.86

No of persons who books and arrive for flight, X is binomial, since there are two outcomes and each person is independent of the other

The probability that if Air-USA books 24 persons, not enough seats will be available

= P(X=23)+P(x=24)

= 0.1315

B. no, it is not low enough

-------------------------------

The prob we got is >10% also

A. no, it is not low enough

7 0
3 years ago
Rewrite the fraction with the denominator of 33 6/11=/33
Sonja [21]
You would time the denominator by 3 so 11*3 would be 33 then times the numerator by the same so 6*3 soot would be 18/33
7 0
2 years ago
4 2/9 × 6 answer with a mixed number in simplest form. :) Thanks!!!
astra-53 [7]
4 2/9=8/9
8/9 x 6=48/9= 5 3/9
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5 0
3 years ago
PLEASE HELP
lisabon 2012 [21]

Step-by-step explanation:

The figure below shows a portion of the graph of the function j\left(x\right) \ = \ 4^{x-2}, hence the average rate of change (slope of the blue line) between the x and x+h is

                     \text{Average rate of change} \ = \ \displaystyle\frac{\Delta y}{\Delta x} \\ \\ \rule{3.7cm}{0cm} = \dsiplaystyle\frac{f\left(x+h\right) \ - \ f\left(x\right)}{\left(x \ + \ h \right) \ - \ x} \\ \\ \\  \rule{3.7cm}{0cm} = \displaystyle\frac{f\left(x + h\right) \ - \ f\left(x\right)}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x+h-2} \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x-2+h} \ - \ 4^{x-2}}{h}

                                                            \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h}\right) \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h} \ - \ 1 \right)}{h}

7 0
1 year ago
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