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3 years ago

Can someone plz help me on this plz

Mathematics

1 answer:

ddd [48]3 years ago

5 0

Ok so ... the answer is B:264 because you multiply all the numbers because volume = l x w x h... which also is 6 x 11 x 4 and that’s how I got B :)

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A rectangular made of square tiles measures 8 tiles wide and 10 tiles long what is the length in tiles of a similar rectangle 12

Agata [3.3K]

There’s are ratio which would be 8/10. 12/x should equally 8/10. So it would be 14 tiles long.

3 0

3 years ago

If we wish to obtain a 95% confidence interval of a parameter using the bootstrap method, we determine the _______ percentile

BigorU [14]

Answer:

2.5th percentile and the 97.5th percentile.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

So we obtain the 0.025*100 = 2.5th percentile and the (1-0.025)*100 = 97.5th percentile.

So the answer is:

2.5th percentile and the 97.5th percentile.

8 0

3 years ago

There are 4 jacks in a standard deck of 52 playing cards. If Patricia selects a card at random, what is the probability that it

n200080 [17]

Look at the attached picture⤴

Hope it will help you.

7 0

3 years ago

Read 2 more answers

Gas Use

Alisiya [41]

Answer: 10.75
Steps:
After 110miles, it used 2.75gallons
Therefore, the amount of gas used per mile is: 100/2.75= 40 gallon per mile

So after 430miles, the number of gallons used: 430/40 = 10.75 gallons

Hope this helps ʕ•ᴥ•ʔ

7 0

3 years ago

Answer the question in the picture

nekit [7.7K]

Recall the angle sum identities:

$\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

Now,

$\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$

Divide through numerator and denominator by $\cos x\cos y$ to get

$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

Next, we use the fact that $x,y$ lie in the first quadrant to determine that

$\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2$

$\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}$

So we then have

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

$\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1$

Finally,

$\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73$

4 0

3 years ago