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3 years ago

5

Corey got 35% off coupon for HEB the store bill came out to $73.80 how nuch money does he owe?

1 answer:

spin [16.1K]3 years ago

7 0

$73.80 - 35% = $47.97 Corey owes $47.97

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A bar of soap weighs as much as 3/4 of an identical bar plus 3/4 of a pound. How much does the bar of soap weigh?

Advocard [28]

3/4=0.75
Since 3/4 (of the bar) + 1/4 (of the bar) equals a whole bar (1), and 3/4 (of the bar) + 0.75 equals the whole weight of the bar, 0.75 is 1/4 of the bar. 1/4 times 4 equals 1, and 0.75 times 4 equals 3, so the bar of soap weighs 3 pounds.

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3 years ago

What is the distance between the points (-4,0) and (-7, -14) ?

vlada-n [284]

Answer:

Step-by-step explanation:

(-4,0) ; (-7, -14)

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\\\ =\sqrt{(-7-[-4])^{2}+(-14-0)^{2}}\\\\ =\sqrt{(-7+4)^{2}+(-14)^{2}}\\\\=\sqrt{(-3)^{2}+(-14)^{2}}\\\\=\sqrt{9+196} \\\\=\sqrt{205} \\\\=14.3178$

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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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3 years ago

Two fractions equivalent to 5/4

Eduardwww [97]

Answer:

10/8, 20/16

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks

Iteru [2.4K]

Let

$P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}$

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

$P(1):\ 1 = \dfrac{1\cdot 2}{2}=1$

So, the base case is ok. Now, we need to assume $P(n)$ and prove $P(n+1)$ .

$P(n+1)$ states that

$P(n+1):\ 1+2+\ldots+n+(n+1) = \dfrac{(n+1)(n+2)}{2}=\dfrac{n^2+3n+2}{2}$

Since we're assuming $P(n)$ , we can substitute the sum of the first n terms with their expression:

$\underbrace{1+2+\ldots+n}_{P(n)}+n+1 = \dfrac{n(n+1)}{2}+n+1=\dfrac{n(n+1)+2n+2}{2}=\dfrac{n^2+3n+2}{2}$

Which terminates the proof, since we showed that

$P(n+1):\ 1+2+\ldots+n+(n+1) =\dfrac{n^2+3n+2}{2}$

as required

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3 years ago