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3 years ago

An audio cable is 15.9 dm long. How long is the cable in meters?

Mathematics

2 answers:

Leto [7]3 years ago

8 0

Answer:

An audio cable is 15.9 dm long. The length of the audio cable in metre is 159m

Solution:

Given that audio cable is 15.9 dm long. We have to convert 15.9 decimetre to meter.

To convert decimetre(m) to metre(cm) we use the given formula

We know,

1m =10 dm

Therefore to convert any number in decimetre (dm) to metre (m) we multiply the given number by 10.

Here we have to convert 15.9 dm to m. Therefore multiplying 15.9 with 10 we get,

$15.9 \times 10 = 159 m$

Thus the length of the cable in metre is 159 m.

sveta [45]3 years ago

3 0

1.59 meters is the correct answer i think.

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Supposef(0) = 2 and 2 ≤ f '(x) ≤ 7for all x in the interval [−5, 5]. determine the greatest and least possible values of f(2).le

IgorC [24]

If the slope of the function is 2, the amount it can change over the interval 0–2 is

... 2×2 = 4 units

If the slope is 7, the amount it can change over the interval 0–2 is

... 2×7 = 14 units

The least possible value of f(2) is 2+4 = 6.

The greatest possible value of f(2) is 2+14 = 16.

7 0

3 years ago

B -5 ≥ -45 answer and steps

Natalka [10]

Answer:

-50

Step-by-step explanation:

b-5 $\geq -45$

b-5 = -45 ( collect like terms)

b = -45 + -5

b = -45 -5 b = -50

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2 years ago

Maggie is practicing her penalty kicks for her upcoming soccer game. During the practice, she attempts 10 penalty kicks. If each

arlik [135]

Answer:

$P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)$

And we can find the individual probabilities like this:

$P(X=8)=(10C8)(0.65)^0 (1-0.65)^{10-8}=0.176$

$P(X=9)=(10C9)(0.65)^1 (1-0.65)^{10-9}=0.072$

$P(X=10)=(10C10)(0.65)^2 (1-0.65)^{10-10}=0.013$

And adding we got:

$P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)= 0.176+0.072+0.013=0.262$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=10, p=0.65)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Solution to the problem

For this case we want to find this probability:

$P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)$

And we can find the individual probabilities like this:

$P(X=8)=(10C8)(0.65)^0 (1-0.65)^{10-8}=0.176$

$P(X=9)=(10C9)(0.65)^1 (1-0.65)^{10-9}=0.072$

$P(X=10)=(10C10)(0.65)^2 (1-0.65)^{10-10}=0.013$

And adding we got:

$P(X \geq 8)= P(X=8) +P(X=9) +P(X=10)= 0.176+0.072+0.013=0.262$

5 0

3 years ago

I need help with this question plz

pashok25 [27]

First you need to distribute the exponent into the terms for the first parenthesis.

(-3x²y³)³ = (-3)³ * (x²)³ * (y³)³

= -27 *x⁶ *y⁹

= -27x⁶y⁹

Then you multiply -27x⁶y⁹ by x²y.

You just need to add the exponents of x and y:

-27x⁶⁺²y⁹⁺¹

= -27x⁸y¹⁰

5 0

3 years ago

A triangle can be formed with side lengths 2in, 3in and 6 in? true or false?

Aleks04 [339]

Answer:

No, a triangle cannot be constructed with sides of 2 in., 3 in., and 6 in.

For three line segments to be able to form any triangle you must be able to take any two sides, add their length and this sum be greater than the remaining side.

in.

For a triangle with sides 3 in., 4 in. and 5 in. which can form a triangle:

3 + 4 = 7 which is greater than 5

3 + 5 = 8 which is greater than 4

4 + 5 = 9 which is greater than 3

Step-by-step explanation:

4 0

3 years ago

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