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2 years ago

A picture measuring 2.5" high by 6" wide is to be enlarged so that the width is now 18 inches. How tall will the picture be?

1 answer:

8 0

We have that
the picture original is 2.5 in x 6 in
high=2.5 in
wide=6 in

the picture <span>enlarged is
</span>high=2.5*x
wide=6*x-----> 6*x=18--------> x=18/6----------> x=3
x is the factor scale
so
high=2.5*x-----> 2.5*3----> 7.5 in

the answer is
7.5 in

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Trevor is analyzing a circle, y2 + x2 = 100, and a linear function g(x). Will they intersect?

lakkis [162]

Given a circle described by the equation:
$y^2+x^2=100$
and a function g(x) given by the table
$\begin{center}
\begin{tabular}{ |c |c |c }
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 -1 & -22 \\
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1 & -18 
\end{tabular}
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The function g(x) describes a straight line with the equation:
$\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1} \\ \\ \frac{y-(-22)}{x-(-1)} = \frac{-20-(-22)}{0-(-1)} \\ \\ \frac{y+22}{x+1} = \frac{-20+22}{0+1} = \frac{2}{1} \\ \\ y+22=2(x+1)=2x+2 \\ \\ y=2x-20$

To check if the circle and the line intersects, we substitute the equation of the line into the equation of the circle to see if we have a real solution.
i.e.
$y^2+x^2=100 \\ \\ (2x-20)^2+x^2=100 \\ \\ 4x^2-80x+400+x^2=100 \\ \\ 5x^2-80x+300=0 \\ \\ x^2-16x+60=0 \\ \\ (x-6)(x-10)=0 \\ \\ x-6=0 \ or \ x-10=0 \\ \\ x=6 \ or \ x=10$

When x = 6, y = 2(6) - 20 = 12 - 20 = -8 and when x = 10, y = 2(10) - 20 = 20 - 20 = 0

Therefore, the circle and the line intersect at the points (6, -8) and (10, 0).

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2 years ago

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Each inch on a map of Roswell, New Mexico represents $0.51\text{ mile}$. If two hills in Roswell are $7$ inches apart on the map

pav-90 [236]

Answer:

$3.57$

Step-by-step explanation:

we know that

The scale factor on a map is equal to

$\frac{1}{0.51}\frac{inch}{miles}$

by proportion

Find how many miles are in the real life $7$ inches apart on the map

$\frac{1}{0.51}\frac{inch}{miles}=\frac{7}{x}\frac{inches}{miles}\\ \\x=0.51*7\\ \\x=3.57 \ miles$

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2 years ago

What are the coordinates of vertex J of the pre-image?(14 POINTS. PLS HELP, GEOMETRY)

DaniilM [7]

If we look at where it is now, we would have multiplied the x and y values by 1/2. since we are working backwards, lets multiply by 2 instead...giving us the vertex for the pre-image....

J is now (0, -2) multiply by 2
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In a certain community, eight percent of all adults over age 50 have diabetes. If a health service in this community correctly d

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Complete question is;

In a certain community, 8% of all people above 50 years of age have diabetes. A health service in this community correctly diagnoses 95% of all person with diabetes as having the disease, and incorrectly diagnoses 10% of all person without diabetes as having the disease. Find the probability that a person randomly selected from among all people of age above 50 and diagnosed by the health service as having diabetes actually has the disease.

Answer:

P(has diabetes | positive) = 0.442

Step-by-step explanation:

Probability of having diabetes and being positive is;

P(positive & has diabetes) = P(has diabetes) × P(positive | has diabetes)

We are told 8% or 0.08 have diabetes and there's a correct diagnosis of 95% of all the persons with diabetes having the disease.

Thus;

P(positive & has diabetes) = 0.08 × 0.95 = 0.076

P(negative & has diabetes) = P(has diabetes) × (1 –P(positive | has diabetes)) = 0.08 × (1 - 0.95)

P(negative & has diabetes) = 0.004

P(positive & no diabetes) = P(no diabetes) × P(positive | no diabetes)

We are told that there is an incorrect diagnoses of 10% of all persons without diabetes as having the disease

Thus;

P(positive & no diabetes) = 0.92 × 0.1 = 0.092

P(negative &no diabetes) =P(no diabetes) × (1 –P(positive | no diabetes)) = 0.92 × (1 - 0.1)

P(negative &no diabetes) = 0.828

Probability that a person selected having diabetes actually has the disease is;

P(has diabetes | positive) =P(positive & has diabetes) / P(positive)

P(positive) = 0.08 + P(positive & no diabetes)

P(positive) = 0.08 + 0.092 = 0.172

P(has diabetes | positive) = 0.076/0.172 = 0.442

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2 years ago