Ten increased by the quotient of a number and 6 is 5

A. 60 degrees B. 120 degrees C. 30 degrees D. 150 degrees

kondaur [170]

Answer:

yu

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Determine the geometric mean of 45 and 15

Alenkinab [10]

I believe 30 is the answer :)

7 0

2 years ago

Read 2 more answers

Which data set could be represented by the box plot shown below?

Fudgin [204]

Answer:

Any set of data that satisfies the 5-Number summary: 1,6,12,16 and 19 can be represented with the box plot.

Step-by-step explanation:

Interpreting Box Plots

A box plot is used to present the 5-Number summary of a set of data.

The 5-Number summary consists of the following in their order of appearance on the box plot.

Minimum Value
First Quartile, $Q_1$
Median, $Q_2$
Third Quartile, $Q_3$
Maximum Value

In the box plot, the following rules applies

The whisker starts from the minimum value and ends at the first quartile.

The box starts at the first quartile and ends at the third quartile. There is a vertical line inside the box which shows the median.
The end whisker starts at the third quartile and ends at the maximum value.

Using these, we interpret the given box plot

A left whisker extends from 1 to 6.

Minimum Value=1
First Quartile =6

The box extends from 6 to 16 and is divided into 2 parts by a vertical line segment at 12.

Median=12
Thrid Quartile=16

The right whisker extends from 16 to 19.

Maximum Value =19

Therefore any set of data that satisfies the 5-Number summary: 1,6,12,16 and 19 can be represented with the box plot.

6 0

3 years ago

On a recent trip to a local orchard, the Morgan family picked four different kinds of apples - Braeburn, Cortland, Fuji, and Rom

ivanzaharov [21]

Answer:

Morgan family picked 144 Braeburn apples, 96 Cortland apples, 72 Fuji apples and 48 Rome apples.

Step-by-step explanation:

Let B, C, F and R represent Braeburn apples, Cortland apples, Fuji apples, and Rome apples respectively.

We have been given that Morgan family picked a total of 360 apples. We can represent this information in an equation as:

$B+C+F+R=360...(1)$

We are told that Morgan family picked twice as many Braeburn as Fuji. We can represent this information in an equation as:

$B=2F...(2)$

We are told that Morgan family picked twice as many Cortland as Rome, that is:

$C=2R...(3)$

We are told that Morgan family picked 50% more Fuji than Rome, that is:

$F=1.5R...(4)$

Upon substituting equation (4) in equation (2), we will get:

$B=2(1.5)R$

$B=3R$

Now, each apple in in terms of Rome apples, so we will get:

$3R+2R+1.5R+R=360$

Let us solve for R.

$7.5R=360$

$\frac{7.5R}{7.5}=\frac{360}{7.5}$

$R=48$

Therefore, Morgan family picked 48 Rome apples.

Upon substituting $R=48$ in (3), we will get:

$C=2R\Rightarrow 2(48)=96$

Therefore, Morgan family picked 96 Cortland apples.

Upon substituting $R=48$ in (4), we will get:

$F=1.5R\Rightarrow 1.5(48)=72$

Therefore, Morgan family picked 72 Fuji apples.

Upon substituting $F=72$ in (2), we will get:

$B=2F\Rightarrow 2(72)=144$

Therefore, Morgan family picked 144 Braeburn apples.

3 0

3 years ago

State the number of possible triangles that can be formed using the given measurements.

romanna [79]

Answer: 39) 1 40) 2

41) 1 42) 0

Step-by-step explanation:

39) ∠A = ? ∠B = ? ∠C = 129°

a = ? b = 15 c = 45

Use Law of Sines to find ∠B:

$\dfrac{\sin B}{b}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin B}{15}=\dfrac{\sin 129}{45}\rightarrow \quad \angle B=15^o\quad or \quad \angle B=165^o$

If ∠B = 15°, then ∠A = 180° - (15° + 129°) = 36°

If ∠B = 165°, then ∠A = 180° - (165° + 129°) = -114°

Since ∠A cannot be negative then ∠B ≠ 165°

∠A = 36° ∠B = 15° ∠C = 129° is the only valid solution.

40) ∠A = 16° ∠B = ? ∠C = ?

a = 15 b = ? c = 19

Use Law of Sines to find ∠C:

$\dfrac{\sin A}{a}=\dfrac{\sin C}{c} \rightarrow\quad \dfrac{\sin 16}{15}=\dfrac{\sin C}{19}\rightarrow \quad \angle C=20^o\quad or \quad \angle C=160^o$

If ∠C = 20°, then ∠B = 180° - (16° + 20°) = 144°

If ∠C = 160°, then ∠B = 180° - (16° + 160°) = 4°

Both result with ∠B as a positive number so both are valid solutions.

Solution 1: ∠A = 16° ∠B = 144° ∠C = 20°

Solution 2: ∠A = 16° ∠B = 4° ∠C = 160°

41) ∠A = ? ∠B = 75° ∠C = ?

a = 7 b = 30 c = ?

Use Law of Sines to find ∠A:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{7}=\dfrac{\sin 75}{30}\rightarrow \quad \angle A=13^o\quad or \quad \angle A=167^o$

If ∠A = 13°, then ∠C = 180° - (13° + 75°) = 92°

If ∠A = 167°, then ∠C = 180° - (167° + 75°) = -62°

Since ∠C cannot be negative then ∠A ≠ 167°

∠A = 13° ∠B = 75° ∠C = 92° is the only valid solution.

42) ∠A = ? ∠B = 119° ∠C = ?

a = 34 b = 34 c = ?

Use Law of Sines to find ∠A:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b} \rightarrow\quad \dfrac{\sin A}{34}=\dfrac{\sin 119}{34}\rightarrow \quad \angle A=61^o\quad or \quad \angle A=119^o$

If ∠A = 61°, then ∠C = 180° - (61° + 119°) = 0°

If ∠A = 119°, then ∠C = 180° - (119° + 119°) = -58°

Since ∠C cannot be zero or negative then ∠A ≠ 61° and ∠A ≠ 119°

There are no valid solutions.

6 0

2 years ago