All categories

3 years ago

PLEASE ANSWER I NEED HELP!!..

1 answer:

5 0

Please don't post masses of questions all at once. 1-2 problems per post would be more realistic and fairer to others who need help.

Regarding <span>x-2/5 = 1/5

We are to solve for x. To do this, let's remove the fractions by mult. all 3 terms by 5. Then 5x - 2 = 1, or 5x = 3.

Dividing both sides by x, we get x = 3/5 (answer). Check this by subst. 3/5 into the original equation.</span>

You might be interested in

From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ

bulgar [2K]

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$ where $C(n,x)=\frac{n!}{x!(n-x)!}$
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
$=0.027$

(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

8 0

3 years ago

What’s the length of the mid segment of a trapezoid with bases of length 16 and 24

blagie [28]

Answer:

Step-by-step explanation:

The midsegment of a trapezoid is calculated as

$\frac{sumofbases}{2}$ = $\frac{16+24}{2}$ = $\frac{40}{2}$ = 20

7 0

3 years ago

Which translation vectors could have been used for the pair of figures?

Ierofanga [76]

if set up like

1 2

3 4

5 6

it would be 4 and 6

7 0

3 years ago

Read 2 more answers

If 14 g of a radioactive substance are present initially and 3 yr later only 7 g remain, how much of the substance will be pres

marishachu [46]

It would be 48 bc if you add it all up and subtract it and add again and multiply and divide and multiply within a and the substance of present after 6 yr you will see that i have wasted your time and that is not the answer

5 0

2 years ago

Use a net to find the surface Area of the square pyramid The height of the pyramid is 9 yes and the base sides are all 2.5 yrds

Cerrena [4.2K]

Answer:

$Area = 51.68yd^2$