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3 years ago

Is 9/10 larger than 13/14

Mathematics

1 answer:

Reptile [31]3 years ago

5 0

Answer:

13/14 is greater

Step-by-step explanation:

Find the lcm for both:

9 x 7 = 63

10 x 7 70

13 x 5 = 65

14 x 5 70

so therefore 13/14 is greater since it has a greater value when finding the lcm

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In a box there are 6 orange balls, 4 green balls and "x" blue balls. If the probability of removing a blue ball is 1/3, what is

worty [1.4K]

Answer: 2/3 I think

Step-by-step explanation:

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2 years ago

Maksim231197 [3]

Answer:

D. x=1

Step-by-step explanation:

Image can be rewritten as 3x + 6 = 9.

Solve for x.

3x +6 = 9

Isolate x by subtracting 6 from both sides.

3x = 3

Get rid of the 3 by dividing by 3 on both sides.

x = 1

Hope this helped.

8 0

2 years ago

You are knitting a blanket. You want the area of the blanket to be 24 ft². You want the length of the blanket to be 2 feet longe

Finger [1]

With the facts you got here, you can make the following equation:

x*(x+2)=24

x^2+2x=24
x^2+2x-24=0

Then you can use the following formula (which you might have seen before):
x = (-b +/- sqrt(b^2-4ac))/2a

where
ax^2+bx+c=0

(-2 +/- sqrt(4+96))/2
(-2 +/- 10)/2

x_1 = 4
x_2 = -6

3 0

3 years ago

Name the pair of vertical angles

Anit [1.1K]

ABC , dce is the answer

7 0

3 years ago

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is:

$74.2-(1.8331)\frac{5.3083}{\sqrt{10} }$ ≤ μ ≤ $74.2+(1.8331)\frac{5.3083}{\sqrt{10} }$

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0

3 years ago