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3 years ago

7

Help me on question 4 plssssssssssss its science not maths

1 answer:

olga55 [171]3 years ago

5 0

Answer:

idk

Step-by-step explanation:

You might be interested in

Which of these is true and why

LuckyWell [14K]

Answer:

Step-by-step explanation:

True statements

All equallateral triangles are similar. Their sides are all in the same ratio when comparred.

All squares are similar. Same reason as equilateral triangles. All sides to both squares compared are the same.

False Statements

Isosceles triangles can and usually do have different base angles.

rectangles can have all sorts or side lengths. The only requirement is consecutive sides form right angles.

2 rhombuses can have side lengths that are in the same ratio, but the heights are not in the same ratio. That eleminates.

Answer

These are the only true ones: Statements 2 and 5 are true. The rest are not.

7 0

1 year ago

Fittoniya [83]

Answer:

Step-by-step explanation

5 0

2 years ago

Read 2 more answers

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

3 years ago

Solve x(x-1) (X-5) <=O

lara [203]

Answer:

x <= 0 or 1 = < x <= 5.

Step-by-step explanation:

First we find the critical points:

x(x - 1)(x - 5) = 0

gives x = 0, x = 1 and x = 5.

Construct a Table of values:

<u> x < 0 </u> <u>x = 0 </u> 0<u>< x < 1</u> <u>1 =< x <= 5</u> <u>x = 5</u>

x <0 0 >0 <0 0

x - 1 <0 -1 >0 <0 0

x - 5 < 0 0 > 0 <0 0

x(x-1)(x-5) < 0 0 >0 <0 0

So the answers are x =< 0 or 1 =< x <= 5.

8 0

3 years ago

Kade has 100 flag pins. He has 221 frog pins and 10 dinosaur pins. How many flag and frog pins does Kade have?

bazaltina [42]

Answer:

321

Step-by-step explanation:

because 100+221 equals 321

Hope this helped.

5 0

3 years ago