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JulsSmile [24]
3 years ago
15

Find the GCF of 30 * 3 and 12 * 4

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

3


Step-by-step explanation:


wlad13 [49]3 years ago
5 0
30:15x2                 12:1x12
                                    2x6                         
                                    3x4           GCF:3
     3x10
     1x30
     5x6
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The useful life of a radial tire is normally distributed with a mean of 30,000 miles and a standard deviation of 5000 miles. The
otez555 [7]

Answer:

  empirical rule: 81.5%

  table or calculator: 81.9%

Step-by-step explanation:

The lower limit of the life range of interest has a z-score of ...

  z = (x -μ)/σ = (20,000 -30,000)/5,000 = -2

The upper limit has a z-score of ...

  z = (35,000 -30,000)/5,000 = 1

<u>Empirical rule solution</u>

The empirical rule tells you that 95% of the distribution lies within 2 standard deviations of the mean, so (100% -95%)/2 = 2.5% lie below z = -2. It also tells you 68% lie within 1 standard deviation of the mean, so (100% -68%)/2 = 16% lie above z = 1.

The fraction that lies within -2 to 1 standard deviations of the means is thus ...

  (100% -2.5% -16%) = 81.5%

The probability the tire has a life in the desired range is about 81.5%.

__

<u>Calculator solution</u>

A probability calculator for the Normal distribution tells you that ...

  P(-2 < z < 1) ≈ 0.8185946

The probability the tire has a life in the desired range is about 81.9%.

6 0
2 years ago
This Venn diagram shows the pizza topping preferences for 9 students. Let event A = The student likes pepperoni. Let event B = T
Lelechka [254]

Answer:

P(A\ or\ B)=\frac{7}{9}

Step-by-step explanation:

We need to use the formula to calculate the probability of (A or B) where  

A=Probability a student likes pepperoni

B=Probability a student likes olive

A and B =Probability a student likes both toppings in a pizza

A or B =Probability a student likes pepperoni or olive (and maybe both), a non-exclusive or

The formula is

P(A\ or\ B)=P(A)+P(B)-P(A\ and\ B)

Since 6 students like pepperoni out of 9:

P(A) = \frac{6}{9}

Since 4 students like olive out of 9:

P(B) = \frac{4}{9}

Since 3 students like both toppings out of 9

P(A\ and\ B) = \frac{3}{9}

Then we have

P(A\ or\ B)=\frac{6}{9}+\frac{4}{9}-\frac{3}{9}

P(A\ or\ B)=\frac{7}{9}

6 0
2 years ago
Simple Interest Problem<br><br> Suppose 30=a×60.<br> What is 30/a
alexandr1967 [171]
The answer is 60 i think
3 0
2 years ago
Read 2 more answers
What is the greatest common factor of 14 and 6?
sergij07 [2.7K]

Answer:

gcf=2

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
99 POINT QUESTION, PLUS BRAINLIEST!!!
Elden [556K]
We draw region ABC. Lines that connect y = 0 and y = x³ are vertical so:
(i) prependicular to the axis x - disc method;
(ii) parallel to the axis y - shell method;
(iii) parallel to the line x = 18 - shell method.

Limits of integration for x are easy x₁ = 0 and x₂ = 9.
Now, we have all information, so we could calculate volume.

(i)

V=\pi\cdot\int\limits_a^bf^2(x)\, dx\qquad\implies \qquad a=0\qquad b=9\qquad f(x)=x^3


V=\pi\cdot\int\limits_0^9(x^3)^2\, dx=\pi\cdot\int\limits_0^9x^6\, dx=\pi\cdot\left[\dfrac{x^7}{7}\right]_0^9=\pi\cdot\left(\dfrac{9^7}{7}-\dfrac{0^7}{7}\right)=\dfrac{9^7}{7}\pi=\\\\\\=\boxed{\dfrac{4782969}{7}\pi}

Answer B. or D.

(ii)

V=2\pi\cdot\int\limits_a^bx\cdot f(x)\, dx


V=2\pi\cdot\int\limits_0^{9}(x\cdot x^3)\, dx=2\pi\cdot\int\limits_0^{9}x^4\, dx=&#10;2\pi\cdot\left[\dfrac{x^5}{5}\right]_0^9=2\pi\cdot\left(\dfrac{9^5}{5}-\dfrac{0^5}{5}\right)=\\\\\\=2\pi\cdot\dfrac{9^5}{5}=\boxed{\dfrac{118098}{5}\pi}

So we know that the correct answer is D.

(iii)
Line x = h

V=2\pi\cdot\int\limits_a^b(h-x)\cdot f(x)\, dx\qquad\implies\qquad h=18


V=2\pi\cdot\int\limits_0^9\big((18-x)\cdot x^3\big)\, dx=2\pi\cdot\int\limits_0^9(18x^3-x^4)\, dx=\\\\\\=2\pi\cdot\left(\int\limits_0^918x^3\, dx-\int\limits_0^9x^4\, dx\right)=2\pi\cdot\left(18\int\limits_0^9x^3\, dx-\int\limits_0^9x^4\, dx\right)=\\\\\\=2\pi\cdot\left(18\left[\dfrac{x^4}{4}\right]_0^9-\left[\dfrac{x^5}{5}\right]_0^9\right)=2\pi\cdot\Biggl(18\biggl(\dfrac{9^4}{4}-\dfrac{0^4}{4}\biggr)-\biggl(\dfrac{9^5}{5}-\dfrac{0^5}{5}\biggr)\Biggr)=\\\\\\

=2\pi\cdot\left(18\cdot\dfrac{9^4}{4}-\dfrac{9^5}{5}\right)=2\pi\cdot\dfrac{177147}{10}=\boxed{\dfrac{177147\pi}{5}}

Answer D. just as before.

6 0
3 years ago
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