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3 years ago

Round 569.903507815 to the nearest whole number?

Mathematics

1 answer:

Monica [59]3 years ago

3 0

Answer:

if you round it to a whole number... it should be 570

Step-by-step explanation:

just round the tenth place

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Solve the following system of equations by Graphing. Label the solution and all intercepts. 4x-2y=8 y=1/2x+2

vovangra [49]

Answer:

The points that intercepts are (4,4) that's what I got. I hope that helps you.

7 0

2 years ago

Find the values for a and b that would make the equality true.

Crank

Answer:

a = - 4, b = 5

Step-by-step explanation:

Expand the left side and compare the coefficients of like terms on the right side.

- 3(2x² + ax + b)

= - 6x² - 3ax - 3b

Comparing like terms with - 6x² + 12x - 15

x - term → - 3a = 12 ( divide both sides by - 3 )

a = - 4

constant term → - 3b = - 15 ( divide both sides by - 3 )

b = 5

6 0

3 years ago

Please help, today's due. Maths lesson is postulate.

zhenek [66]

You need to put everything to one side and equal to 120. Then you combine all like terms.
then you subtract 24 with 120. You get 94/24=4

10x+6x+16+8x+8=120
24x+24=120
-24 -24
____________
24x=96
/24 /24
X=4

8 0

2 years ago

Which expression can be used to find the area of the square? Question 4 options: 2(6.4+6.4) 6.4² 6.4×2 2(6.4×6.4)

Nimfa-mama [501]

Option 2 => 6.4^2
Because area of a square = side *side or side^2
Hope this helps u..!!!

6 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

2 years ago