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3 years ago

Solving a Multi-Step Equation

Mathematics

2 answers:

hodyreva [135]3 years ago

7 0

Your answer will be 15 no need to thank me

VLD [36.1K]3 years ago

5 0

Answer:

x = 15

Step-by-step explanation:

I have this all put under a photo. Let me know if you need any explanation

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Part A picture above

-BARSIC- [3]

Answer:

Part A C = (10,5) Part B C. D'(0,10)

Step-by-step explanation:

Part A

Since c is at the point (2,1) in relation to the origin, we can multiply those distances by our scale factor of 5

(2,1) * 5 = (10,5)

The new point C is going to be (10,5)

Part B

If you dilate with a factor of 5 -- relative to the origin -- you have to multiply the distance from the origin by 5.

In this case, point D is already on the y axis, so it's x value wouldn't be affected. Point D is currently 2 units away from (0,0), so we can multiply 2*5 to get 10 -- our ending point is (0,10)

8 0

3 years ago

Maya needs $58 to go on a field trip. She has saved $18.50. She earns $6.50 per hour cleaning her neighbor's garden, and she ear

insens350 [35]

She has $18.50 so she needs $39.50.

6.50×3= 19.50

18.50+19.50= 38

5.25×4=21

38+21=59

Maya would have enough money to go on the trip.

6 0

3 years ago

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

3 years ago

Which mathematical expression means seven more than the product of three and x

vladimir2022 [97]

Answer:

More than means addition and product means multiplication. Therefore, seven more than the product of three and x would mean multiplying 3 and x together then adding 7 to that. Or, 3x + 7.

Step-by-step explanation:

7 0

3 years ago

What is the volume of the composite figure shown? (Use 3.14 for π.)

Nimfa-mama [501]

$The\ volume\ of\ the\ cone:V_1=\frac{1}{3}\pi r^2 H\\(r-a\ radius;\ H-height)\\------------------\\r=7ft;\ H=7ft\\\\V_1=\frac{1}{3}\pi\cdot7^2\cdot7=\frac{1}{3}\pi\cdot343=\frac{343}{3}\pi\ (ft^2)\\-------------------\\The\ volume\ of\ the\ cylinder:V_2=\pi r^2 H\\(r-a\ radius;\ H-height)\\-------------------\\r=7ft;\ H=6ft\\\\V_2=\pi\cdot7^2\cdot6=294\pi\ (ft^3)\\------------------------\\The\ volume\ of\ the\ composite\ figure: V_F=V_1+V_2$

$V_F=\frac{343}{3}\pi+294\pi=\frac{343}{3}\pi+\frac{294\cdot3}{3}\pi=\frac{343}{3}\pi+\frac{882}{3}\pi=\boxed{\frac{1225}{3}\pi\ (ft^3)}\\\\\approx\frac{1225}{3}\cdot3.14}=\frac{3846.5}{3}\approx\boxed{1,282.17\ (ft^3)}\leftarrow answer$

7 0

4 years ago