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love history [14]
3 years ago
7

Algebra factorize a4 - 3a2b2 + b4​

Mathematics
1 answer:
Vesnalui [34]3 years ago
8 0

Answer:

The term a⁴-3a²b²+b⁴ can't be factorised

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A store has a 40\%40%40, percent off sale on headphones. With this discount, the price of one pair of headphones is \$36$36dolla
12345 [234]
Put it in an equation;
it is given that 36 is the price after the 40% discount, so;
36=X*40%
X here is the original price
36=X*(40/100)
40X=3600
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6 0
3 years ago
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The length of a phone conversation is normally distributed with a mean of 4 minutes and a standard deviation of .6 minutes. What
Ivanshal [37]

Answer:

0.04746

Step-by-step explanation:

To answer this one needs to find the area under the standard normal curve to the left of 5 minutes when the mean is 4 minutes and the std. dev. is 0.6 minutes.  Either use a table of z-scores or a calculator with probability distribution functions.

In this case I will use my old Texas Instruments TI-83 calculator.  I select the normalcdf( function and type in the following arguments:  :

normalcdf(-100, 5, 4, 0.6).  The result is 0.952.  This is the area under the curve to the left of x = 5.  But we are interested in finding the probability that a conversation lasts longer than 5 minutes.  To find this, subtract 0.952 from 1.000:   0.048.  This is the area under the curve to the RIGHT of x = 5.

This 0.048 is closest to the first answer choice:  0.04746.

8 0
3 years ago
At carl's combine diner, there are three size of coffee drinks regular (300ml), large (500ml) and extra large (800mL), and they
MariettaO [177]

Answer:

The number of regular, large, and extra-large drinks are 12, 15, and 10 respectively.

Step-by-step explanation:

Given that the cost for regular coffee drinks (300 ml)=$2.25

The cost for large coffee drinks (500 ml)=$3.25

The cost for extra large coffee drinks (800 ml)=$5.75

Let p,q, and r be the number of regular, large, and extra-large coffee sold.

As the diner sold a total of 37 coffees, so

p+q+r=37

r=37-p-q...(i)

The volume of p regular coffee = 300p ml

The volume of q  large coffee = 500q ml

The volume of r extra-large coffee = 800r ml

As the total volume of coffee sold was 19,100mi, so

300p+500q+800r=19100

By using equation (i)

300p+500q+800(37-p-q)=19100

300p+500q+800 x 37 - 800p - 800q=19100

-500p-300q=19100-29600

500p+300q=10500

500p=10500-300q

p=21-0.6q...(ii)

Now, the cost of p regular coffee=$2.25p

The cost of q large coffee=$3.25q

The cost of r extra-large coffee=$5.75r

As the amount of money made in coffee sales was $133.25, so

2.25p+3.25q+5.75r=133.25

By using equations (1)  we have

2.25p+3.25q+5.75(37-p-q)=133.25

2.25p+3.25q+212.75-5.75p-5.75q=133.25

3.50p+2.50q=79.5

From equation (ii)

3.5(21-0.6q)+2.50q=79.5

73.5-2.1q+2.5q=79.5

0.4q=79.5-73.5=6

q=6/0.4

q=15

From equation (ii)

p=21-0.6(15)

p=12

From equation (i)

r= 37-12-15

r=10

Hence, the number of regular, large, and extra-large drinks are 12, 15, and 10 respectively.

5 0
3 years ago
A construction company can remove
kolbaska11 [484]

Answer:

sorry not having idea about ans

6 0
3 years ago
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Which expression is equivalent to −4(2x − 8)−2x ?
Vesnalui [34]
C. -10x+32

Explanation: Solve the problem
( -4(2x-8)-2x )
5 0
2 years ago
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