All categories

3 years ago

Which statement best explains the relationship between numbers divisible by 5

Mathematics

1 answer:

frosja888 [35]3 years ago

8 0

Answer:

a number that is divisible by 10 is also divisible by 5 since all factor of 10 are also factor of 5

You might be interested in

A dressmaker needs to cut 12-inch pieces of ribbon from rolls of ribbon that are 9 feet in length. How many 12-inch pieces can t

Tems11 [23]

Answer:

so the question is very interesting and the answer is 12 × 15 so the answer is 180 rolls of ribbons

8 0

3 years ago

Mark made a scale drawing of a soccer field, using a scale of .5 cm=1m. The actual length of the field is 110 m. What is the len

nataly862011 [7]

Answer:

The length of the field on the drawing is 55 cm.

Step-by-step explanation:

Given:

Mark made a scale drawing of a soccer field.

Using a scale of .5 cm=1 m.

The actual length of the field is 110 m.

Now, to find the length of the field on drawing.

Let the length of the field on drawing be $x.$

As given 0.5 cm is equivalent to 1 m.

Thus, $x$ is equivalent to 110 m.

Now, to get the length of the field on drawing by using cross multiplication method:

$\frac{0.5}{1} =\frac{x}{110}$

<em>By cross multiplying we get:</em>

⇒ $55=x$

⇒ $x=55$

Therefore, the length of the field on the drawing is 55 cm.

8 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

Solve the system of equations y=x-4, y=-3x

abruzzese [7]

Y=x-4
x=y+4
y=-3(y+4)
y=-3y-12
4y=-12
y=-3
x=1

5 0

3 years ago

Yeah I need help…..<br> Do your work soldiers

Aloiza [94]

I can’t, sir! I cannot leave my post to help with your math problems, over!

8 0

2 years ago

Read 2 more answers