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2 years ago

Find the area of the rhombus

Mathematics

2 answers:

nalin [4]2 years ago

5 0

The answer is 108 .

professor190 [17]2 years ago

3 0

Answer:

108

Step-by-step explanation:

none. I looked it up...so ion know lol. Im just tryna get these 50 points for sum reason...

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The dimensions of a rectangle can be given by x+7 and x+2. If the area of the rectangle is 66 square inches, what are the dimens

Nadusha1986 [10]

to solve for the dimensions (x+7)(x+2)=66,

we can first use the foiling method to simplify the left side.

x^2 + 2x + 7x + 14 = 66

x^2 + 9x + 14 = 66

now, subtract 66 from both sides.

x^2 + 9x - 52 = 0

now, split this into two parentheses.

(x + 13)(x - 4)

since the root of -13 would give you negative values, x=4. This means that the dimensions of the rectangle are 11 and 6.

7 0

2 years ago

Add 3 to h, then subtract 9 from the result

Anon25 [30]

Answer:

3h-9

Step-by-step explanation:

you can't fit it into an equation because 1, i dont a image of it and 2, you can't subtract a number from a coefficient/variable

7 0

3 years ago

-3a(a? – 10a + 25) =

yan [13]

a
2
+
10
a
+
25
Rewrite
25
25
as
5
2
5
2
.
a
2
+
10
a
+
5
2
a
2
+
10
a
+
5
2
Check the middle term by multiplying
2
a
b
2
a
b
and compare this result with the middle term in the original expression.
2
a
b
=
2
⋅
a
⋅
5

4 0

3 years ago

Read 2 more answers

The distance between flaws on a long cable is exponentially distributed with mean 12 m.

Elden [556K]

Answer:

(a) The probability that the distance between two flaws is greater than 15 m is 0.2865.

(b) The probability that the distance between two flaws is between 8 and 20 m is 0.3246.

(d) The standard deviation is 12.

(e) The 65th percentile of the distances is 12.61 m.

Step-by-step explanation:

The random variable X can be defined as the distance between flaws on a long cable.

The random variable X is exponentially distributed with mean, μ = 12 m.

The parameter of the exponential distribution is:

$\lambda=\frac{1}{\mu}=\frac{1}{12}=0.0833$

The probability density function of X is:

$f_{X}(x)=0.0833e^{-0.0833x};\ x\geq 0$

(a)

Compute the probability that the distance between two flaws is greater than 15 m as follows:

$P(X\geq15)=\int\limits^{\infty}_{15}{0.0833e^{-0.0833x}}\, dx\\=0.0833\times \int\limits^{\infty}_{15}{e^{-0.0833x}}\, dx\\=0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{\infty}_{15}\\=e^{0.0833\times 15}\\=0.2865$

Thus, the probability that the distance between two flaws is greater than 15 m is 0.2865.

(b)

Compute the probability that the distance between two flaws is between 8 and 20 m as follows:

$P(8\leq X\leq20)=\int\limits^{20}_{8}{0.0833e^{-0.0833x}}\, dx\\=0.0833\times \int\limits^{20}_{8}{e^{-0.0833x}}\, dx\\=0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{20}_{8}\\=e^{0.0833\times 8}-e^{0.0833\times 20}\\=0.51355-0.1890\\=0.32455\\\approx0.3246$

Thus, the probability that the distance between two flaws is between 8 and 20 m is 0.3246.

(c)

The median of an Exponential distribution is given by:

$Median=\frac{\ln (2)}{\lambda}$

Compute the median as follows:

$Median=\frac{\ln (2)}{\lambda}$

$=\farc{0.69315}{0.08333}\\=8.322$

Thus, the median is 8.322.

(d)

The standard deviation of an Exponential distribution is given by:

$\sigma=\sqrt{\frac{1}{\lambda^{2}}}$

Compute the standard deviation as follows:

$\sigma=\sqrt{\frac{1}{\lambda^{2}}}$

$=\sqrt{\frac{1}{0.0833^{2}}}\\=12.0048\\\approx 12$

Thus, the standard deviation is 12.

(e)

Let x be 65th percentile of the distances.

Then, P (X < x) = 0.65.

Compute the value of x as follows:

$\int\limits^{x}_{0}{0.0833e^{-0.0833x}}\, dx=0.65\\0.0833\times \int\limits^{x}_{0}{e^{-0.0833x}}\, dx=0.65\\0.0833\times |\frac{e^{-0.0833x}}{-0.0833}|^{x}_{0}=0.65\\-e^{-0.0833x}+1=0.65\\-e^{-0.0833x}=-0.35\\-0.0833x=-1.05\\x=12.61$

Thus, the 65th percentile of the distances is 12.61 m.

4 0

2 years ago

Between which pair of numbers is the exact product of 379 and8?

wariber [46]

3032 this is the answer

3 0

2 years ago

Read 2 more answers