Answer:
16% of its popular porcelain tile will have breaking strengths greater than 412.5 pounds per square inch.
Step-by-step explanation:
We are given that the breaking strength of its most popular porcelain tile is normally distributed with a mean of 400 pounds per square inch and a the standard deviation of 12.5 pounds per square inch.
Let X = <u><em>the breaking strength of its most popular porcelain tile</em></u>
SO, X ~ Normal()
The z score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = mean breaking strength of porcelain tile = 400 pounds per square inch
= standard deviation = 12.5 pounds per square inch
Now, probability that the popular porcelain tile will have breaking strengths greater than 412.5 pounds per square inch is given by = P(X > 412.5)
P(X > 412.5) = P( > ) = P(Z > 1) = 1 - P(Z 1)
= 1 - 0.84 = <u>0.16</u>
Therefore, 16% of its popular porcelain tile will have breaking strengths greater than 412.5 pounds per square inch.
Answer:
B
Step-by-step explanation:
A linear association means the function is a relation that can be shown with a line.
The first scatterplot is in the form of a parabola, and parabolas are the quadratic function form, meaning it cannot be linear.
Next, the second scatterplot goes diagonally, in a straight line. If we were to connect the dots using a line, then it would be a linear association, meaning that B is the correct answer.
Answer:
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Step-by-step explanation:
For the answer to the question above, <span>f x is the number of days she works, she'll earn $90x </span>
<span>After buying the laptop, she'll have $90x - $700 left over, which will pay for ($90x - $700) / $150 days of travel. So we have y = ($90x - $700) / $150 = (9x - 70) / 15 = 0.6x - (14/3) </span>
<span>Note that y can't be negative. Also, if y = 0, then Emma doesn't get to travel at all, so we should avoid that. So we have: </span>
<span>0.6x - (14/3) > 0 </span>
<span>0.6x > 14/3 </span>
<span>x > (14/3) / 0.6 </span>
<span>x > 70/9 </span>
<span>The question says that x can be up to 40, so the domain is 70/9 < x <= 40 </span>
<span>That's approximately 7.777... < x <= 40 </span>
<span>Multiply those numbers by 0.6 and then subtract 700 to get the range: </span>
<span>0 < y <= 58/3 </span>
<span>That's approximately 0 < y <= 19.333</span>