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3 years ago

Suppose x=9 and y=3 what is x when y=21

Mathematics

1 answer:

madreJ [45]3 years ago

6 0

Answer = 63

Explanation

You could set up a ratio of y1/y2 = x1/x2
In your case

3/21=9/x
3x=189
X=63

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Help a girl out with geometry

Darya [45]

Follow the steps

106=x+79+x+45
106=2x+124
-124= -124
-18=2x
( -18)/2=(2x)/2 <----- solving for x
-9=x

then
angle LMF= -9+79
=70

angle FMN= -9+45
= 36
(when you add 70 and 36 together you get 106 which is the measurement of angle LMN or in other word the whole thing)

In conclusion the answer to the problem is:
Angle LMF is equal to 70.

7 0

3 years ago

The area of a square rug is 64 ft.² what is the perimeter of the rug?

Slav-nsk [51]

Answer:

$32$ ${ft}^{2}$

Step-by-step explanation:

The area of a square rug = 64 ft²

First we will find the side of square rug = ${64}^{2} = {s}^{2}$

${s}^{2} = 8$

Perimeter of the rug = $4 \times 8$ $= 32$ ft²

<h3>Hope it is helpful...</h3>

4 0

2 years ago

PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!!

yulyashka [42]

Answer:

x=-10, y=-20

Step-by-step explanation:

Since both 2x and 3x+10 equals y, they equal each other.

2x = 3x + 10

x = -10

Substitute -10 for x

2(-10) = y

y=-20

7 0

3 years ago

Suppose that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in t

hram777 [196]

Answer:

a. $\mu=15$

b. $\mu=7.8586 \ and \ \mu=22.1414$

c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Step-by-step explanation:

a.Binomial distribution is defined by the expression

$P(X=k)=C_k^n.p^k.(1-p)^{n-k}$

Let n be the number of trials, $n=100$

and p be the probability of success, $p=15\%$

The mean of a binomial distribution is the probability x sample size.

$\mu=np=100\times0.15=15$

b.Limits within which p is approximately 95%

sd of a binomial distribution is given as: $\sigma=\sqrt npq\\q=1-p$

Therefore, $\sigma=\sqrt(100\times0.015\times0.85)=3.5707$

Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.

$sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414$

Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).

c. $x=45$ is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

4 0

3 years ago

Let n be a natural number. Show that 3 | n3 −n

Vladimir [108]

Let $n=1$ . Then $n^3-n=1^3-1=0$ . By convention, every non-zero integer $n$ divides 0, so $3\vert n^3-n$ .

Suppose this relation holds for $n=k$ , i.e. $3\vert k^3-k$ . We then hope to show it must also hold for $n=k+1$ .

You have

$(k+1)^3-(k+1)=(k^3+3k^2+3k+1)-(k+1)=(k^3-k)+3(k^2+k)$

We assumed that $3\vert k^3-k$ , and it's clear that $3\vert 3(k^2+k)$ because $3(k^2+k)$ is a multiple of 3. This means the remainder upon divides $(k+1)^3-(k+1)$ must be 0, and therefore the relation holds for $n=k+1$ . This proves the statement.

4 0

3 years ago