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3 years ago

Use the order of operations to evaluate this expression (-2+1)

Mathematics

2 answers:

timama [110]3 years ago

5 0

Answer:

${( - 2 + 1)}^{2} + 5(12 \div 3) - 9 \\ 2 + 1 + 5 \times 4 - 9 \\ 3 + 20 - 9 \\ 23 - 9 \\ 14$

strojnjashka [21]3 years ago

3 0

Answer:

Step-by-step explanation:

2²×3

I hope its correct

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Isolated triangle 68 perimeter equal sides more than 10 ft use x for the shorter side formula

Ugo [173]

Answer:

Step-by-step explanation:

Find the perimeter of an isosceles triangle whose equal sides have a size of 10 m each and the angle between them equal to 30°. We need to know all sides in order to find the perimeter of this triangle. Let x be the base of this isosceles triangle.

6 0

2 years ago

What two numbers multiply to equal -27 but adds up to equal -6

kkurt [141]

Answer:

-9 and 3

Step-by-step explanation:

as -9×3= -27

and

-9+3= -6

4 0

3 years ago

Solve -2x + 13 = -7x + 28

SpyIntel [72]

Answer:

The answer would be 46x

Step-by-step explanation:

-2x+13=11

-7x+28=35

46x

8 0

3 years ago

A student solved the equation 2x+6=12 using algebra tiles. incorrectly says the solution is 9. Solve the equation. What mistake

alexdok [17]

Instead of subtracting 6 from 12 then dividing they added 6 to 12 & ended up with 18. 18 by 2 is 9.

8 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