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2 years ago

Find the value of x that makes p||q

Mathematics

1 answer:

padilas [110]2 years ago

4 0

Answer:

1. x = 30

2. x = 39

Step-by-step explanation:

4x and ( 3x + 30 ) are alternate exterior angles,

Alternate exterior angles are equal.

4x = 3x + 30

4x - 3x = 30

x = 30

x and ( 3x + 24 ) are co - interior angles.

Sum of co - interior angles is 180,

x + 3x + 24 = 180

4x + 24 = 180

4x = 180 - 24

4x = 156

x = 156 / 4

x = 39

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Jeff has a density of 1000 g/mL and a volume of 25-mL. What is Jeff’s mass?

zlopas [31]

Answer:

25,000 grams

Step-by-step explanation:

Density can be found using the following formula:

$d=\frac{m}{v}$

If we rearrange the formula for $m$ , mass, we will multiply both sides of the equation by v.

$d*v=\frac{m}{v}*v$

$d*v=m$

The mass can be found by multiplying the density by the volume.

Jeff's density is 1000 grams per milliliter and his volume is 25 milliliters.

$d= 1000 g/mL\\v=25 mL$

Substitute the values into the formula.

$m=d*v$

$m= 1000 g/mL * 25 mL$

Multiply. When you multiply, the "mL" or milliliters will cancel out.

$m= 1000 g * 25$

$m= 25000 g$

Jeff's mass is 25,000 grams.

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

Will someone please check my answers asap?? will give a brainliest

Usimov [2.4K]

Your choices are correct.

1. adding -3 to the function shifts its graph downward by 3 units.

2. 7^(2x) = (7^2)^x = 49^x

3. f(0) = -6; g(0) = -3, f(0) = 2×g(0)

3 0

3 years ago

HELP PLEASE! 15 PTS

defon

The answer is 7/10 because the even values are 2,4,6, and the others are 7,8,9,10. Added together, it adds up to 7/10

4 0

3 years ago

Read 2 more answers

A certificate of deposit often charges a penalty for withdrawing funds before the maturity date. If the penalty involves two mon

Aleonysh [2.5K]

Answer:

The amount for early withdrawal is $204

Step-by-step explanation:

Annual interest rate = 5% = 0.05

Investment = $24,000

Annual interest rate * Investment = Annual interest

0.05 * $24,000 = $1,200

Penalty = Two months of interest

(months/months in a year) * Annual interest

(2/12) * $1,200 = 0,17

0,17 * $1,200 = $204,00

6 0

2 years ago