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3 years ago

6

Evaluating expressions

1 answer:

Dvinal [7]3 years ago

5 0

Answer:

There is no expression to evaluate.

Step-by-step explanation:

You might be interested in

What is 5% added to $194?

leva [86]

Answer:

203.7

Step-by-step explanation:

5% of 194 added to 194 =

= 5% * 194 + 194

= 0.05 * 194 + 194

= 9.7 + 194

= 203.7

6 0

3 years ago

Please help for both of these questions! its due soon! i'll mark brainlist

Vsevolod [243]

For the first question the answer is
X=-9 or X=10

First you factor the left side of the question
(X+9)(X-10)=0 then u set factors equal to zero
X+9=0 or x -10=0
X=-9 or X=10
Now for question two is the second bubble of answer

7 0

3 years ago

Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

What is the best approximation of the area of a circle with a diameter of 17 meters? Use 3.14 to approximate pi. 53.4 m² 106.8 m

Arte-miy333 [17]

R = 17/2 =8.5 m
A = 3.14 x (8.5)^2
A = 3.14 x 72.25
A = 226.865

answer <span>226.9 m² (third choice)</span>

8 0

3 years ago

Reduce the fraction -20t^5 u^2 v^3 over 48t^7 u^4 v

timofeeve [1]

.........................................................

5 0

3 years ago