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3 years ago

Can someone help me, please!!!

Mathematics

1 answer:

kolezko [41]3 years ago

7 0

Answer:

The line shows an intercept of -3 and a slope of 5/7. Equation D displays this

Step-by-step explanation:

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Can someone tell me the answer to this question please?????

katrin [286]

Answer:

C) MK

Step-by-step explanation:

You only follow the letters as their order

In first triangle, so VX = MK

3 0

3 years ago

In a genetics class, 6 students have GREEN eyes, 5 students have BLUE eyes, and 9 have HAZEL eyes. If a single student is picked

mixer [17]

EXPLANATION:

Given;

We are given that in a class there are the following groups of students;

$\begin{gathered} Green\text{ }eyes=6 \\ Blue\text{ }eyes=5 \\ Hazel\text{ }eyes=9 \end{gathered}$

Required;

We are required to calculate the probability that a student selected at random will have Green eyes OR Blue eyes.

Step-by-step solution;

To calculate the probability of an event, we shall use the following formula;

$P[Event]=\frac{Number\text{ }of\text{ }required\text{ }outcomes}{Number\text{ }of\text{ }all\text{ }possible\text{ }outcomes}$

To calculate the probability that a selected student will have green eyes;

$P[green]=\frac{6}{20}=\frac{3}{10}$

To calculate the probability that a selected student will have blue eyes;

$P[blue]=\frac{5}{20}=\frac{1}{4}$

The probability of event A or event B will be the addition of probabilities.

Therefore, the probability that a randomly selected student will have green or blue eyes will be;

$P[G]+P[B]=\frac{3}{10}+\frac{1}{4}$ $P[F]+P[B]=\frac{6}{20}+\frac{5}{20}=\frac{11}{20}$

Therefore,

ANSWER:

$P[G\text{ }orB]=\frac{11}{20}$

6 0

1 year ago

A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

mr Goodwill [35]

Answer: 8.26 m

Step-by-step explanation:

$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$

$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$

$$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}$

$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$

$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$