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2 years ago

Find the square root of 312.5

Mathematics

2 answers:

Vadim26 [7]2 years ago

6 0

Answer:

The square root is 17.6776695297

arsen [322]2 years ago

4 0

Answer:

Changes made to your input should not affect the solution:

(1): "f3" was replaced by "f^3".

Unauthorized use of the imaginary unit "i" or syntax error in complex arithmetic expression

.... V

find^2sqrtof312.5

The symbol "i" is only allowed in complex arithmetic, for example:

(3/5+7i)+(0.3-7.002i)

(5+77i)-(19/4-8i)

(240-22i)*(247/7+2.222i)

(33/5+99i)/(33/5-88i)

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A(n)=−6−4(n−1)<br> Find the 4th term in the sequence

blondinia [14]

If n = the term number, then:
A(4) = 6-4(4-1)
A(4) = 6-4(3)
A(4) = 6-12
A(4) = -6
A = -1.5

4 0

3 years ago

Please help! Thank you so much!

igor_vitrenko [27]

Angle W:

180 = 90 + 57 + W

180 = 147 + W

W = 33 degrees

Using trigonometry functions to find the side lengths. (SOH CAH TOA)

Side XZ:

cos(57) = XZ / 18

XZ = cos(57) x 18

XZ = 9.8 units

Side XW:

sin(57) = XW / 18

XW = sin(57) x 18

XW = 15.1 units

Hope this helps!! :)

4 0

3 years ago

A submarine descended 3 feet per minute.If the surface of the water is 0, which integer describes the position of the submarine

marin [14]

Answer:

-24

Step-by-step explanation:

-3*8=-24

8 0

3 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

5 0

3 years ago

Is 1.375 =to 2lbs i need help with this

erastovalidia [21]

Answer:

Step-by-step explanation:

5 0

3 years ago

Find the square root of 312.5​

Find the square root of 312.5