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juin [17]
3 years ago
13

Help me please thank you

Mathematics
1 answer:
grin007 [14]3 years ago
5 0
2 3/4 - 1 = 1 3/4 or 1.75
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On moving day, Jorge needs to rent a truck. The length of the cargo space is 12 ft, and the height is 1 ft less than the width.
lidiya [134]

Answer:

12ft \times7ft \times6ft

Step-by-step explanation:

Given that truck can hold = 504 ft^{3} i.e.

Volume, V = 504 ft^{3}

Length of cargo space = 12 ft

Let width of cargo space = w ft

As per question statement:

Let height of cargo space = (w-1) ft

To find: The Dimensions of Cargo Space

Formula for Volume of Cargo Space:

V = Length \times Width \times Height

Putting the given values and conditions:

504 = 12 \times w \times (w-1)

\Rightarrow w(w-1) = \dfrac{504}{12}\\\Rightarrow w^{2} -w = 42\\\Rightarrow w^{2} -w - 42=0\\\Rightarrow w^{2} -7w +6w -42 =0\\\Rightarrow w(w -7) +6(w -7) =0\\\Rightarrow (w -7)(w+6) = 0\\\Rightarrow w =7, -6

Dimensions can not be negative, so width, w = 7 ft

Height = (w-1) = 7-1 = 6 ft

So, the dimensions are 12ft \times7ft \times6ft.

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3 years ago
How do u divide fractions
8090 [49]
For example 5/10\1/2 you would keep the 5/10 and flip the 1/2 to 2/1 and then multiply 5\10x2/1 to get 10/10 so 1
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the tickets are 90 dollars in all adults 7 kids 12 kids are 2/3 cost of adults tickets. What is the cost of the adult tickets
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I hope it helps. .........

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Find the percent of increase from 15 to 20
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Read 2 more answers
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

0.95 probaiblity of hitting a target

This means that p = 0.95

10 targets

This means that n = 10

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

P(X = 10) + P(X < 10) = 1

We want P(X < 10). So

P(X < 10) = 1 - P(X = 10)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987

P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401

40.1% probability that he will miss at least one of them

7 0
3 years ago
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