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yulyashka [42]
2 years ago
12

What value of x makes this equation true?

Mathematics
1 answer:
Softa [21]2 years ago
3 0
Well you gotta find out what x means
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What is the equation of a line that has a slope of 0 and goes through the point (4, 10)?
kompoz [17]

Answer:

y =10

Step-by-step explanation:

The equation of a line in point slope form is expressed as;

y - y0 = m(x-x0)

m is the slope

(x0, y0) is the point on the line

Given

m = 0 and (x0, y0) = (4, 10)

On substituting;

y - 10 = 0(x-4)

y - 10 = 0

y = 0+10

y = 10

Hence the required equation of the line is y =10

4 0
3 years ago
Evaluate a+b/2,if we know a=3 and b =6
Crazy boy [7]
4.5. because ⁼3=9, and 9 divided by 2 is 4.5
5 0
3 years ago
Read 2 more answers
MATH!!!! HELP ME!!!!!! PLEASE!!!!!
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6 0
3 years ago
Find the equation of a line that is perpendicular line to y = 10x - 45 and goes through (1, 1).
Airida [17]

The line y = 10x - 45 is already written in the y = mx+q form. This means that m is the slope. If two lines are perpendicular, their slopes are the anti-inverse of each other, i.e. their product is -1.

So, our perpendicular line has a slope of -\frac{1}{10}

Finally, we want a line passing through (1,1) with slope -\frac{1}{10}:

y-1 = -\dfrac{1}{10}(x-1) \iff y = -\dfrac{x}{10} + \dfrac{1}{10}+1 = -\dfrac{x}{10} + \dfrac{11}{10}

7 0
3 years ago
A cookie factory monitored the number of broken cookies per pack yesterday.
trapecia [35]

Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given  

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20  

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

sqrt(37.69) = 6.139

s*sqrt(n-1) = 18.2

s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } } \leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }

confidence interval:

(18.2/7.948) < S < (18.2/6.139)

2.290 < S < 2.965

8 0
3 years ago
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