Lucy receives 28 dollars a night for babysitting. She also had 135 dollars in

Umnica [9.8K]

Answer:

I think the figure of N and E of 48in is 2304. The figure of I and E and N and A of 32in is 1024. The figure of L and I and L and A is 144.

Step-by-step explanation:

To find the area, you have to multiply of each side in each figure. For example, Figure A and B and C and D of 16in is 16 × 16 which is 256.

A = Area

a = Side

A = 2304, A = a^2 = 48^2 = 2304.

A = 1024, A = a^2 = 32^2 = 1024.

A = 144, A = a^2 = 12^2 = 144.

I hope this answers your question!

8 0

2 years ago

Do you know how to use the Pythagorean identity?

goblinko [34]

The Pythagorean theorem or identity is given by

H²=P²+B²
P²=H²-B²
B²=H²-P²

Where

B is base of right angle traingle
H is Hypotenuse
P is perpendicular

8 0

1 year ago

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

Likurg_2 [28]

Answer:

(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

$M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}$

$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

Hence, (a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$ .

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

6 0

3 years ago

Someone please help??

irga5000 [103]

Answer:

Im not 100% sure but i can tell you it is (D)

Step-by-step explanation:

8 0

2 years ago

What tells us how many steady beats go in a measure?

Step2247 [10]

Answer:

You'll spot the time signature in the beginning of the music – it's two numbers stacked vertically. The top number tells you how many beats there are in one measure.

Step-by-step explanation:

hope it helps

5 0

1 year ago