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kherson [118]
3 years ago
14

2,688÷96= what? Please answer at the bottom

Mathematics
2 answers:
pickupchik [31]3 years ago
7 0

Answer:

28

Step-by-step explanation:

Yep

marishachu [46]3 years ago
5 0

Answer:

28 is the correct answer

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Multiply using the rule for the square of a binomial.
GalinKa [24]
To solve:
(x-10)(x-10)

Answer: x^2-20x+100
5 0
3 years ago
I walk each day to school. By what percentage would I need to increase my usual average speed in order for the journey to take 2
lesya692 [45]

Answer:

25%

Step-by-step explanation:

Let v, t be the current average speed and the time taken to reach the school respectively.

As distance = speed x time, so,

distance, d=vt...(i)

Let V be the new average speed in order to to take 20% less time than t.

So, time taken with speed V = t-20% of t = t- 0.2t= 0.8t

As distance is constant, so

d= V(0.8t)= 0.8Vt

hBy using equation (i), we have

0.8Vt = vt

0.8 V = v

V= v/0.8=1.25v

Therefore, the percentage increase in the average speed = \frac{V-v}{v}\times 100

=\frac{1.25v-v}{v}\times 100

=25%

Hence, the percentage increase in the average speed is 25%.

8 0
3 years ago
A model satellite has a scale of 1 in: 6 ft.
insens350 [35]

Answer:

B) 18 ft

Step-by-step explanation:

The scale is 1 in. model : 6 ft real.

6/1 = 6

To go from model size to real size, multiply the model size by 6 and change from inches to feet.

3 in. model = 3 × 6 ft = 18 ft

5 0
1 year ago
Just say 1,2,3,4,5,6 or 7 and there are more than 1 answer WORTH 90 POINTS!! HELP ME​
alisha [4.7K]
Yo bro i got to add words but the answer is 2
7 0
3 years ago
A tobacco company claims that the amount of nicotine in its cigarettes is a randomvariable with mean 2.2 mg and standard deviati
Genrish500 [490]

Answer:

0.0000

Unusual

Step-by-step explanation:

Given that a tobacco company claims that the amount of nicotine in its cigarettes is a random variable with mean 2.2 mg and standard deviation .3 mg.

i.e. population parameters are

\mu =2.2 \\s = 0.3

The approximate probability that the sample meanwould have been as high or higher than 3.1

=P(X\geq 3.1)\\=P(Z\geq \frac{3.1-2.2}{\frac{0.3}{\sqrt{100} } } )\\=P(Z\geq 30)\\

=0.0000

8 0
4 years ago
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