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3 years ago

Solve following equation for number x.verify that your solution is correct -15=8x+1

Mathematics

1 answer:

GalinKa [24]3 years ago

3 0

-15x=8x+1
-8
-23x=1
÷-23
X= -1/23
Check: -15 (-1/23)=8 (-1/23)+1
15/23=-8/23+1
15/23=15/23 correct

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40% of the groups budget was spent on activities. If $64,000 was spent on the activities, what was the full budget? solve with t

Andre45 [30]

Answer:

The full budget has a total value of $160,000

Step-by-step explanation:

This is a simple ratio problem and can be solved using the Rule of Three property. The Rule of Three is used to compare two ratios and find the missing part. In this case the ratios would be the following.

$\frac{64,000}{40 percent} = \frac{x}{100 percent}$

$\frac{64,000 * 100 percent}{40 percent} = \frac{x}$

$\frac{6,400,000}{40 percent} = \frac{x}$

$160,000= \frac{x}$

So now we know that 100% (The full budget) has a total value of $160,000.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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3 years ago

The table below shows some inputs and outputs of the invertible function f with domain all real numbers.

Ratling [72]

Answer:

${f}^{ - 1} (f(93)) = 93$

$f(f(7)) = - 15$

Step-by-step explanation:

A function and its inverse has the following properties

${f}^{ - 1} (f(x)) = x$

This implies that;

${f}^{ - 1} (f(93)) = 93$

From the table,

$f(7) = - 6$

This means that:

$f(f(7)) = f( - 6)$

$f(f(7)) = - 15$

Note that: f(7) means the value of f(x) when x=7.

From the table f(x)=-6 when x=7.

That is why f(7)=-6.

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3 years ago

Thetime(inhours)requiredtorepairamachineis an exponentially distributed random variable with parameter λ = 1 . What is 2 (a) the

mihalych1998 [28]

Answer:

a) 0.1353

b) 0.3679

Step-by-step explanation:

Let's start by defining the random variable T.

T : ''The time (in hours) required to repair a machine''

T ~ exp (λ)

T ~ exp (1)

The probability density function for the exponential distribution is

(In the equation I replaced λ = L)

$f(x)=Le^{-Lx}$

With L > 0 and x ≥ 0

In this exercise λ = 1 ⇒

$f(x)=e^{-x}$

For a)

$P(T>2)$

$P(T>2)=1-P(T\leq 2)$

$P(T>2)=1-\int\limits^2_0 {e^{-x} } \, dx$

$P(T>2)=1-(-e^{-2}+1)$

$P(T>2)=e^{-2}=0.1353$

For b)

$P(T\geq 10/T>9)$

The event (T ≥ 10 / T > 9) is equivalent to the event T ≥ 1 so they have the same probability of occur

$P(T\geq 10/T>9)=P(T\geq 1)$

$P(T\geq 1)=1-P(T$

$P(T\geq 1)=1-(-e^{-1}+1)=e^{-1}=0.3679$

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