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S_A_V [24]
3 years ago
9

Could somebody help.​

Mathematics
1 answer:
Marianna [84]3 years ago
8 0
The answer is 4/9.

To find the answer you would need to find the values of x and y. Using the given information you know x/5 is equal to 2/3. Then you would cross multiply and divide in order to fine the value of x, which is 10/3. Doing the same thing to find y, which would be 15/2. Since it’s day the ratio of x and y. You would divide the value of x by y to get your answer of 4/9.
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Solve for x<br><br> 7/8 of x is 14
SVEN [57.7K]

7÷8 of x = 14

  1. 8/7×7/8 of x=14×8/7
  2. x=16
8 0
1 year ago
Read 2 more answers
Whats 87 less than the quotient of an unknown number and 43 is -75?
erastovalidia [21]
<span>87 less than the quotient of an unknown number and 43 is -75
Converting that to mathematical equation:
87 - (x / 43) = -75

Evaluating the equation, we can the value of x.
87 - (x/43) = -75
-x/43 = -75 - 87
-x/43 = -162
-x = -6966
x = 6966</span>
4 0
3 years ago
The top and bottom margins of a poster 66 cm each, and the side margins are 44 cm each. If the area of the printed material on t
jasenka [17]

Answer:

  • width: 24 cm
  • height: 36 cm

Step-by-step explanation:

When margins are involved, the smallest area will be the one that has its dimensions in the same proportion as the margins. If x is the "multiplier", the dimensions of the printed area are ...

  (4x)(6x) = 384 cm^2

  x^2 = 16 cm^2 . . . . . divide by 24

  x = 4 cm

The printed area is 4x by 6x, so is 16 cm by 24 cm. With the margins added, the smallest poster will be ...

  24 cm by 36 cm

_____

<em>Comment on margins</em>

It should be obvious that if both side margins are 4 cm, then the width of the poster is 8 cm more than the printed width. Similarly, the 6 cm top and bottom margins make the height of the poster 12 cm more than the height of the printed area.

_____

<em>Alternate solution</em>

Let w represent the width of the printed area. Then the printed height is 384/w, and the total poster area is ...

  A = (w+8)(384/w +12) = 384 +12w +3072/w +96

Differentiating with respect to w gives ...

  A' = 12 -3072/w^2

Setting this to zero and solving for w gives ...

  w = √(3072/12) = 16 . . . . matches above solution.

__

<em>Generic solution</em>

If we let s and t represent the side and top margins, and we use "a" for the printed area, then the above equation becomes the symbolic equation ...

  A = (w +s)(a/w +t)

  A' = t - sa/w^2

For A' = 0, ...

  w = √(sa/t)

and the height is ...

  a/w = a/√(sa/t) = √(ta/s)

Then the ratio of width to height is ...

  w/(a/w) = w^2/a = (sa/t)/a

  width/height = s/t . . . . . . the premise we started with, above

6 0
3 years ago
There are 70 students in the school band. 40% of them are sixth graders, 20% are seventh graders, and the rest are eighth grader
guajiro [1.7K]

Answer:

Sixth graders: 14

Seventh graders: 28

Eighth graders: 28

Step-by-step explanation:

To solve this you just have to use a rule of three to solve the percentages, remember that the 100% will be the 70 students, and we solve each case separetly:

70 students= 100%

sixth grades=   40%

Sixth grades= (40*70)/100

Sixth graders= 28

70 students= 100%

seventh grades=   20%

seventh graders=   (20*70)/100

Seventh graders=14

So if we have that the rest of the students are eighth graders, we just add up the sixth and seventh graders and withdraw them from the total:

28+14=42

70-42=28

SOwe have that the eighth graders are 28

7 0
3 years ago
Use the table of systolic blood pressures (in mm Hg) in each arm
e-lub [12.9K]

Answer:

The 90% confidence interval would be given by (63.330;81.070)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let put some notation

x=left arm , y = right arm

x: 175 169 182 146 144

y: 102 101 94 79 79

Since the Confidence is 0.90 or 90%, the value of a = 0.1 and a/2= 0.05we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,4)".And we see that ta/2= 2.13

3 0
2 years ago
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