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3 years ago

Help me please i dont understand

Mathematics

1 answer:

kipiarov [429]3 years ago

5 0

Answer:

C=75.36

A=452.16

Step-by-step explanation:

C=2x3.14xr

$2$ × $\pi$ × $12=75.36$

A=3.14× $r^{2}$

3.14x12x12=452.16

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Complete the equation that represents the relationship between t and p

fredd [130]

Answer:

p=9*t

Step-by-step explanation:

1*9=9

2*9=18

3*9=27

4*9=36

3 0

3 years ago

A certain standardized test's math scores have a bell-shaped distribution with a mean of 530 and a standard deviation of 119. Co

kumpel [21]

Answer:

a) 68.2%

b) 31.8%

c) 2.3%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 530

Standard Deviation, σ = 119

We are given that the distribution of math scores is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(test scores is between 411 and 649)

$P(411 \leq x \leq 649)\\\\= P(\displaystyle\frac{411 - 530}{119} \leq z \leq \displaystyle\frac{649-530}{119})\\\\= P(-1 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -1)\\= 0.841 - 0.159 = 0.682 = 68.2\%$

b) P(scores is less than 411 or greater than 649)

$P(x < 411 < x, x > 649)\\=1 = P(411 \leq x \leq 649)\\=1 - 0.682\\=0.318 = 31.8\%$

c) P(score greater than 768)

P(x > 768)

$P( x > 768) = P( z > \displaystyle\frac{768 - 530}{119}) = P(z > 2)$

$= 1 - P(z \leq 2)$

Calculation the value from standard normal z table, we have,

$P(x > 768) = 1 -0.977 = 0.023 = 2.3\%$

5 0

4 years ago

Please help me !!!

kati45 [8]

$23)~75r^2-48=3(25r^2-16)=3(5r-4)(5r+4)\\\\
24)~45x^2-5=5(9x^2-1)=5(3x-1)(3x+1)\\\\
25)~m^3n-mn^3=mn(m^2-n^2)=mn(m-n)(m+n)\\\\
26)~6a^3-24ab^2=6a(a^2-4b^2)=6a(a-2b)(a+2b)\\\\
27)~3x^2+15x-72=3(x^2+5x-24)=3(x-3)(x+8)\\\\
28)~2m^2-4m+2=2(m^2-2m+1)=2(m-1)^2\\\\
29)~x^3+9x^2-52x=x(x^2+9x-52)=x(x-4)(x+13)\\\\
30)~4y^3+4y^2-120y=4y(y^2+y-30)=4y(y-5)(y+6)$

5 0

3 years ago

Read 2 more answers

There was no actually pyramid for these questions it tells you the measurements in the problem though. please help me

Nimfa-mama [501]

V=448 in^3
l=8 in
w=12
h=?
V=l•w•h/3
448=8•12•h/3 multiply both sides by 3
3•448=96•h
1344=96•h
h=1344/96
h=14 in

V=l•w•h/3
h=4in
l=3in
w=2.5 in
V=?
V=4•3•2.5/3
V=30/3
V=10 in^3

3 0

3 years ago

How many arrangements can be made using 2 letters of the word HYPERBOLAS if no letter is to be used more than once?

alisha [4.7K]

Answer:

90 arrangements

Step-by-step explanation:

Since there are no repititions of letters, there are unique 10 letters in total.

THe number of arrangements would be 2 permutation 10. We need the formula for permutation. That is:

$nPr=\frac{n!}{(n-r)!}\\$

Now, n = 10 [total] and r is 2, so we have:

$nPr=\frac{n!}{(n-r)!}\\\\_{10}P_{2}=\frac{10!}{(10-2)!}\\=\frac{10!}{8!}\\=\frac{10*9*8!}{8!}\\=10*9\\=90$

So, there can be 90 arrangements

3 0

3 years ago