Findes the slope from two points point 1=(4,5) point 2=(6,9)

Ashley earns $5 per hour babysitting. How much will she earn in 3 hours?

fgiga [73]

Ashley will get $15 because it like 3X5=15

7 0

3 years ago

the length of diagonal of a rectangular field is 23.7 m and one of its sides is 18.8 m. find the perimeter of the field.

katen-ka-za [31]

Answer:

Approximately 66.4 Meters

Step-by-step explanation:

So we have a rectangle with a width of 18.8 meters and a diagonal with 23.7 meters. To find the perimeter, we need to find the length first. Since a rectangle has four right angles, we can use the Pythagorean Theorem, where the diagonal is the hypotenuse.

$a^2+b^2=c^2$

Plug in 18.8 for either a or b. Plug in the diagonal 23.7 for c.

$(18.8)^2+b^2=23.7^2\\b^2=23.7^2-18.8^2\\b=\sqrt{23.7^2-18.8^2} \\b\approx14.4 \text{ meters}$

Therefore, the length is 14.4 meters. Now, find the perimeter:

$P=2l+2w\\P=2(14.4)+2(18.8)\\P=66.4\text{ meters}$

7 0

3 years ago

The accompanying data contains the depth (in kilometers) and magnitude, measured using the Richter Scale, of all earthquakes

Sunny_sXe [5.5K]

Answer:

Depth:

μ =20.2025 km

M = 15.625 km

Range = 47.15 km

σ ≈ 15.92 km

Q₁ = 5.7375 km

Q₃ = 34.6675 km

Magnitude:

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

Step-by-step explanation:

The given data are;

Depth ${}$ Magnitude

0.76 ${}$ 0.84

4.93 ${}$ 0.47

8.16 ${}$ 0.35

33.58 ${}$ 1.32

21.2 ${}$ 1.61

35.03 ${}$ 4.57

10.05 ${}$ 5.52

47.91 ${}$ 1.99

For the Depth, we have;

The mean, μ = (0.76+4.93+8.16+33.58+21.2+35.03+10.05+47.91)/8 =20.2025 km

The median, M = The (n + 1)/2th term after arranging the term in increasing order as follows;

0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 , the median is therefore;

(8 + 1)/2th term or the 4.5th term which is 10.05 + (21.2 - 10.05)/2 = 15.625 km

The Range = The highest - The lowest value = 47.91 - 0.76 = 47.15 km

The Standard deviation of, σ, is given as follows;

$\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{N}}$

Where;

$x_i$ = The individual data point = (0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 )

N = The total number of data point = 8

Substituting, (using Microsoft Excel) we get;

$\sigma =\sqrt{\dfrac{\sum \left (x_i-20.2025 \right )^{2} }{8}} \approx 15.92 \ km$

Q₁ = The first quartile = The (n + 1)/4th = term arranged in increasing order

Q₁ = The (8 + 1)/4th term = The 2.25th term = 4.93 + (8.16 - 4.93)×0.25) = 5.7375 km

Q₃ = The first quartile = The 3×(n + 1)/4th = term arranged in increasing order

Q₃ = The 3×(8 + 1)/4th term = The 6.75th term = 33.58 + 3×(35.03 - 33.58)×0.25) = 34.6675 km

For the magnitude, we have, using the same formulas and procedures as above;

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

4 0

3 years ago

What fractional part of one gallon would fit in one quart

babymother [125]

1/4 1 gallon = 4 quarts 1 quart = 0.25 gallon
I hope this helped!

4 0

3 years ago

Read 2 more answers

We did not find results for: A measure of malnutrition, called the pelidisi, varies directly as the cube root of a person's w

asambeis [7]

$\bf \qquad \qquad \textit{double proportional variation}\\\\
\begin{array}{llll}
\textit{\underline{y} varies directly with \underline{x}}\\
\textit{and inversely with \underline{z}}
\end{array}\implies y=\cfrac{kx}{z}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
p=pedalisi\\
w=weight\\
s=\textit{sitting height}
\end{cases}\quad 
\begin{array}{llll}
%pelidisi, varies directly as the cube root of a person's weight in grams and inversely as the person's sitting height in centimeters.
\textit{pelidisi varies directly}\\
\textit{as cube root of weight}\\
\textit{and inversely to }\\
\textit{sitting height}
\end{array}\implies p=\cfrac{k\sqrt[3]{w}}{s}\\\\
-------------------------------$

$\bf \textit{we know that }
\begin{cases}
w=48,820\\
s=78.7\\
p=100
\end{cases}\implies 100=\cfrac{k\sqrt[3]{48820}}{78.7}
\\\\\\
100\cdot 78.7=k\sqrt[3]{48820}\implies \cfrac{7870}{\sqrt[3]{48820}}=k
\\\\\\
thus\qquad \boxed{p=\cfrac{\frac{7870}{\sqrt[3]{48820}}\sqrt[3]{w}}{s}}
\\\\\\
\textit{now, what is \underline{p} when }
\begin{cases}
w=54,688\\
s=72.6
\end{cases}?\implies p=\cfrac{\frac{7870}{\sqrt[3]{48820}}\sqrt[3]{54688}}{72.6}$

now, if that value is less than 100, then the fellow is "undernourished", otherwise, is overfed.

3 0

3 years ago