All categories

2 years ago

Two sides of a triangle measure

Mathematics

2 answers:

Eduardwww [97]2 years ago

8 0

Answer:

Step-by-step explanation:

Svetllana [295]2 years ago

5 0

Answer:

A:4cm

Step-by-step explanation:

You might be interested in

Question

ololo11 [35]

Answer:

correct experiment: viral culture test

Trial: (RCT) random control trial

It causes mild sickness or death

6 0

2 years ago

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question, the median is:

$Median = 22$ and $N = 14$

To calculate the median, we make use of:

$Median = \frac{N + 1}{2}th$

$Median = \frac{14 + 1}{2}th$

$Median = \frac{15}{2}th$

$Median = 7.5th$

This means that, the median is the average of the 7th and 8th items.

The 7th and 8th items are blanks.

However, from the question; the mode is:

$Mode = 18$

Since the values of the box are in increasing order and the average of 18 and 18 do not equal 22 (i.e. the median), then the 7th item is:

$7th = 18$

The 8th item is calculated as thus:

$Median = \frac{1}{2}(7th + 8th)$

$22= \frac{1}{2}(18 + 8th)$

Multiply through by 2

$44 = 18 + 8th$

$8th = 44 - 18$

$8th = 26$

The updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question.

$Mean = 26$

Mean is calculated as:

$Mean = \frac{\sum x}{n}$

So, we have:

$26= \frac{1 + 2nd + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 12th + 58 + 61}{14}$

Collect like terms

$26= \frac{ 2nd + 12th+1 + 4 + 12 + 15 + 18 + 18 + 26 + 29 + 30 + 32 + 58 + 61}{14}$

$26= \frac{ 2nd + 12th+304}{14}$

Multiply through by 14

$14 * 26= 2nd + 12th+304$

$364= 2nd + 12th+304$

This gives:

$2nd + 12th = 364 - 304$

$2nd + 12th = 60$

From the updated box,

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We know that:

The 2nd value can only be either 2 or 3

The 12th value can take any of the range 33 to 57

Of these values, the only possible values of 2nd and 12th that give a sum of 60 are:

$2nd = 3$

$12th = 57$

i.e.

$2nd + 12th = 60$

$3 + 57 = 60$

So, the complete box is:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

6 0

2 years ago

Graph a system of equations to solve log (−5.6x + 1.3) = −1 − x. Round to the nearest tenth.

const2013 [10]

From the least to the greatest, the solutions are x ≈ -2.1 and x ≈ 0.2

the answers are -2.1 and 0.2.

<h3>What is a linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

The question is incomplete.

The complete question is in the picture, please refer to the attached picture.

We have two systems of the equation:

y = log (−5.6x + 1.3)

y = −1 − x

After plotting the above equation on the graph we will see two intersections point:

(-2.12, 1.12) and (0.221, -1.221)

So the least value is x = -2.12 ≈ -2.1

The greatest value is x = 0.221 ≈ 0.2

Thus, from the least to the greatest, the solutions are x ≈ -2.1 and x ≈ 0.2

the answers are -2.1 and 0.2.

Learn more about the linear equation here:

brainly.com/question/11897796

#SPJ1

8 0

2 years ago

Find the zeros of polynomial function and solve polynomials equations. f(x)=24x^3-64x^2-21x+56

8090 [49]

Since this polynomial has 4 terms, factoring by grouping should be the first thing we try here.

So, we have:

$8x^2(3x-8)-7(3x-8)=0 \implies\\ (8x^2-7)(3x-8)=0$

So, we can use ZPP to find out roots:

$3x-8 =0 \implies\\ x=\frac{8}{3}\\ 8x^2-7=0 \implies \\ 8x^2=7 \implies \\ x^2=\frac{7}{8} \implies \\ x=\pm\frac{\sqrt{7}}{\sqrt{{8}}}\\ \text{Rationalizing denominator:}\\ x=\pm\frac{\sqrt{56}}{8}$

So our three roots are:

$x \in \{\frac{8}{3},\frac{\sqrt{56}}{8},-\frac{\sqrt{56}}{8}\}$

8 0

3 years ago

Khan academy Find the median of the data in the dot plot below. chocolate chips

Helga [31]

Answer:

Where's the picture?

Step-by-step explanation:

Can't see it.

6 0

3 years ago