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shepuryov [24]
3 years ago
10

Drag each equation to show if it could be a correct first step to solving the

Mathematics
1 answer:
Masteriza [31]3 years ago
3 0

Answer:

Here you go man

Step-by-step explanation:

Hope this helps :)

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Para embalar una caja se emplea 4,2 m de cinta adhesiva. ?Cuántas cajas se podrán embalar con tres rollos que tienen 3 hm, 7 dam
kompoz [17]

For this case, we perform the conversions:

First roll:

1 \ Hectometer --------> 100 \ meters\\3 \ Hectometer --------> x

x = \frac {3 * 100} {1}\\x = 300 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 300 meters of adhesive tape.

1 -----------> 4.2

c -----------> 300

c = \frac {300 * 1} {4.2}\\c = 71.42857143\\c = 71

You can pack 71 boxes.

Second roll:

1 \ Decametro --------> 10 \ meters\\7 \ Decameter --------> x\\x = \frac {7 * 10} {1}\\x = 70 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 70 meters of adhesive tape.

1 -----------> 4.2

c -----------> 70

c = \frac {70 * 1} {4.2}\\c = 16.66666667\\c = 16

You can pack 16 boxes.

Third roll:

1 -----------> 4.2

c -----------> 50

c = \frac {50 * 1} {4.2}\\c = 11.9047619\\c = 11

You can pack 11 boxes.

Thus, in total you can pack11 + 16 + 71 = 98 \ boxes

Answer:

98 boxes

4 0
4 years ago
A submarine starts at the surface of the Pacific Ocean and descends 60 feet every hour. What integer expresses the submarine's d
LiRa [457]
-60*6 is what, -360 feet
7 0
3 years ago
Read 2 more answers
Estimate the solution to the system of equations. asap, please!! I will mark for brain list
Darya [45]

Answer:

(1 \frac{1}{3}, 2 \frac{1}{3} )

Step-by-step explanation:

Given the 2 equations

7x - y = 7 → (1)

x + 2y = 6 → (2)

Multiplying (1) by 2 and adding to (2) will eliminate the y- term

14x - 2y = 14 → (3)

Add (2) and (3) term by term to eliminate y

15x = 20 ( divide both sides by 15 )

x = \frac{20}{15} = \frac{4}{3} = 1 \frac{1}{3}

Substitute this value of x into either of the 2 equations and solve for y

Substituting in (2)

\frac{4}{3} + 2y = 6

2y = 6 - \frac{4}{3} = \frac{14}{3} ( divide both sides by 2 )

y = \frac{7}{3} = 2 \frac{1}{3}

7 0
3 years ago
Express 12 2/3 in simplest radical form
natta225 [31]
Hope this will help you if I am answering this right.

6 0
3 years ago
The life times of light bulbs produced by a particular manufacturer have a mean of 1,200 hours and a standard deviation of 400 h
Nezavi [6.7K]

Answer:

a. 1200hours

b.  σ² = 400² =160000hour

c. standard error = 133.33

d. for 9 lightbulb it is likely that 9*0.146 = 1.31 bulbs will have lives of fewer than 1,050 hours?

Step-by-step explanation:

mean of 1,200 hours

σ, a standard deviation of 400 hours.

Sample size, N =  nine bulbs,

a) mean of the sample mean lifetime: is given as 1,200 hours

b) variance of the sample mean is the square of the  σ, a standard deviation of 400 hours.

σ² = 400² =160000hours

c) What is the standard error of the sample mean?

The standard deviation of a sampling distribution of mean values is called the standard error of the means,

standard error of the means,  σx = σ √N

The formula for the standard error of the means is true for all values infinite number of sample, N.

σx = σ √N

     =400  √9 = 400/3 =133.3333

d) the probability that, on average, those nine lightbulbs have lives of fewer than 1,050 hours

The area under part of a normal probability curve is  directly proportional to probability and the value is calculated as

z = ( x₁−x) /σ

where z = propability of normal curve

x₁ = variate mean = 1050hours

x = mean of 1200hours

σ = standard deviation = 400

applying the formula,

z= (1050-1200)/400

z = 150/400 =0.375

Using a table of partial areas beneath the standardized normal curve (see Table of normal curve, a z-value of 0.375 corresponds to an area of 0.1460 between the mean value.

Thus the probability of a lightbulbs having lives of fewer than 1,050 hours is 0.1460.

for 9 lightbulb it is likely that 9*0.146 i.e. 1.31 bulbs will have lives of fewer than 1,050 hours?  

8 0
3 years ago
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