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3 years ago

Pinta are usually labeled as

Mathematics

1 answer:

Agata [3.3K]3 years ago

4 0

Hi there!

I'm going to assume you mean't "Pints", which is abbreviated as Pt.

Hope this helped!

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If 2x - 5= 7x + 3, then x = ?

Leona [35]

Answer: -8/5= x

Step-by-step explanation:

2x-5=7x+3

-5=5x+3

-8=5x

-8/5=x

3 0

3 years ago

Read 2 more answers

A car traveled a distance of 126 miles in three hours traveling at the same speed what distance will the car travel in seven hou

Charra [1.4K]

Answer:

the correct answer would be 294 because you would have to divide 126/3 to find out how many miles per hour then multiply it by 7 for 7 hours

hope this help

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Hi!

Elina [12.6K]

Answer:

404.09x cm³ (to 5 s.f.),

where x is the unknown height of the shape.

Step-by-step explanation:

Volume of the shape

= base area × height

= (large semicircle- small semicircle) × thickness of shape

Area of circle= $\pi {r}^{2}$

Area of semicircle= $\frac{1}{2} (\pi)( {r}^{2} )$

Diameter of small circle= 14

Radius= diameter ÷2

radius= 14 ÷2

radius = 7 cm

Area of small semicircle

$= \frac{1}{2} (\pi)( {7}^{2} ) \\ = \frac{49}{2} \pi$

Radius of large semicircle

= 35 ÷2

= 17.5 cm

Area of large semicircle

$= \frac{1}{2} (\pi)( {17.5}^{2} ) \\ = \frac{1225}{8} \pi$

Base area of shape

$= \frac{1225}{8}\pi - \frac{49}{2} \pi \\ = \frac{1029}{8} \pi$

Since the height if the shape is not given, I'm afraid I cant help you find the volume.

Let the height of the shape be x cm.

Volume of shape= $\frac{1029x}{8} \pi \: cm^{3}$

3 0

3 years ago

Let X and Y be two independent random variables following beta distributions Beta(120, 2020).

Alla [95]

With $X,Y\sim\mathrm{Beta}(120,2020)$ , we have identical PDFs

$P(X=x)=\dfrac{x^{119}(1-x)^{2019}}{B(120,2020)}$

for $0$ , and 0 otherwise, where

$B(a,b)=\dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$

Since $X,Y$ are independent, the joint PDF is

$P(X=x,Y=y)=P(X=x)P(Y=y)=\dfrac{(xy)^{119}((1-x)(1-y))^{2019}}{B(120,2020)^2}$

for points $(x,y)$ in the unit square, and 0 otherwise.

1. The distribution is continuous, so $P(X=0.5)=\boxed0$ .

2. $X+3$ is the region in the $x,y$ plane contained within the unit square and above the line $y=\frac{x+3}2$ . This region is empty, because this line lies above the square altogether, so $P(X+3$ .

3. $X>Y$ is the region in the same square below the line $y=x$ . So we have

$P(X>Y)=\displaystyle\int_0^1\int_0^xP(X=x,Y=y)\,\mathrm dy\,\mathrm dx=\boxed{\frac12}$

4 0

3 years ago

Anyone please answer me faster if know what the answer??!?

Ronch [10]

Answer:

1/2 answer

Step-by-step explanation:

the step you have to do this but I tell you the answer if you can't I will tell you do not worry

7 0

3 years ago