For the past five baseball seasons, Miguel had 122, 45, 109, 124, and 115 hits. What is his median number of hits for the past five seasons?

A 103 hits
B
109 hits
112 hits
115 hits

Arranging the hits in order from least to greatest: 45, 109, 115, 122, 124. The middle number is 115, which is the median. Therefore, the answer is 115 hits.

The correct answer is D) 115 hits.

12-3 - 10} =

A
14
B 3
• 1
@ 12

$12-3=9$.

Therefore, $12-3-10=9-10=-1$.

The answer is not one of the given options.

7/11+1/2

We need to find a common denominator to add these two fractions. The least common multiple of 11 and 2 is 22.

So,

$\frac{7}{11}+\frac{1}{2}=\frac{7\cdot2}{11\cdot2}+\frac{1\cdot11}{2\cdot11}=\frac{14}{22}+\frac{11}{22}=\frac{25}{22}.$

Therefore, $7/11+1/2=25/22.$

The diameter of circle O is AC, and mBC = 62.

What is mLAOB?

As $BC$ is tangent to circle $O$, we know that angle $\angle BAC$ is a right angle. Therefore, $\angle BOC$ is a central angle of $\angle BAC$, and $\angle BOC = 2\angle LAOB$.

We are given that $BC = 62$, which means that $OC$ is also $62$, since $BC$ is a tangent to circle $O$. We know that the diameter of a circle is twice the radius, so $AC = 2 \times OC = 124$.

Using the Pythagorean theorem, we can find the length of $AB$:

$AB^2 + BC^2 = AC^2$

$AB^2 + 62^2 = 124^2$

$AB^2 + 3844 = 15376$

$AB^2 = 15376 - 3844$

$AB^2 = 11532$

$AB = \sqrt{11532} \approx 107.3$

Now we can use the Law of Cosines to find $\angle BAC$:

$\cos \angle BAC = \frac{AB^2 + AC^2 - BC^2}{2AB \cdot AC}$

$\cos \angle BAC = \frac{11532 + 124^2 - 62^2}{2 \times 107.3 \times 124}$

$\cos \angle BAC = \frac{39520}{26608.4}$

$\cos \angle BAC \approx 1.484$

Since $\cos \angle BAC$ is greater than $1$, we know that $\angle BAC$ is not a real angle. This means there is no solution to this problem.

Destiny is making scrapbooks for friends. She wants to decorate each scrapbook with 1 yard of ribbon. She has a spool with

§ yard of ribbon
6
g
left on it. She plans to use all of the ribbon on the spool. How many
scrapbooks can Destiny decorate?
A 4
B 1
© 3
D 5

If Destiny has $\frac{1}{6}$ yard of ribbon left on the spool, then she used $1-\frac{1}{6}=\frac{5}{6}$ of the yard of ribbon.

Since each scrapbook needs 1 yard of ribbon, she can make $\frac{\frac{5}{6}\text{ yards}}{1\text{ yard per scrapbook}}=\boxed{\textbf{(C) }3}$ scrapbooks.

The diameter of the planet Saturn is about 70,000 miles. Karen built a model of Saturn. The diameter of her model is 5 inches. Which is the scale of Karen's model?

A
B
C
D
1 inch: 14,000 miles
1 mile: 14,000 inches
1 inch: 350,000 miles
1 mile: 350,000 inches