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3 years ago

Solve the equation. -3x-5=16 What does 'x' equal?

Mathematics

2 answers:

notsponge [240]3 years ago

7 0

Answer:

-7

Step-by-step explanation:

-3x-5=16

-3x=21

x=-7

Alexandra [31]3 years ago

4 0

Answer:

x=-7

Step-by-step explanation:

-3x-5=16 Addition property of Equality

-3x=21 Multiplication Property of Equality

x=-7

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Jocelyn is knitting a blanket she knits 1/4 foot of the scarf in 3/4 of an hour how fast is jocelyn’s knitting speed in feet per

Alchen [17]

Her speed per hour is 1/3 of a foot

4 0

4 years ago

PLEASE HELP ALGEBRA 2 Question is on the photo below

Sedbober [7]

Answer:

The value of f(x) - g(x) is:

f(x) - g(x) = 2x² + 6x - 4

Step-by-step explanation:

Given

f(x) = 3x² + x - 3

g(x) = x² - 5x + 1

To determine

f(x) - g(x)

Using the formula

f(x) - g(x) = 3x² + x - 3 - ( x² - 5x + 1)

= 3x² + x - 3 - x² + 5x - 1

= 2x² + 6x - 4

Therefore, the value of f(x) - g(x) is:

f(x) - g(x) = 2x² + 6x - 4

6 0

3 years ago

Can someone please give me this answer to question 3 please

s344n2d4d5 [400]

Answer:

kurt

Step-by-step explanation:

5 0

3 years ago

Let me know the answer plz

Gala2k [10]

Answer:

The correct answer option is H. 4374.

Step-by-step explanation:

We are given the following geometric sequence and we are to find its 8th term:

$\frac{2}{9}, \frac{2}{3}, 2,...$

Here $a_1=\frac{2}{9}$ and common ratio $(r) = \frac{\frac{2}{3} }{\frac{2}{9} } =3$ .

The formula we will use to find the 10th term is:

nth term = $a_1 \times r^{(n-1)}$

Substituting the values in the formula to get:

10th term = $\frac{2}{9} \times 3^{(10-1)}$

10th term = 4374

7 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

4 years ago