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2 years ago

Gena wants to estimate the quotient of –21.87 divided by 4.79. Which expression shows the estimate using front-end estimation?

Mathematics

2 answers:

choli [55]2 years ago

8 0

He is actually right ! It’s -21/5

allochka39001 [22]2 years ago

7 0

-21/5 is the closest!!!

Hope this helps.

If you want me to explain why, just ask! :)

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Suppose a shipment of 120 electronic components contains 3 defective components. to determine whether the shipment should be ac

Likurg_2 [28]

For this problem, the most accurate is to use combinations

Because the order in which it was selected in the components does not matter to us, we use combinations

Then the combinations are $nC_r = \frac{ n! }{r! (n-r)!}$

n represents the amount of things you can choose and choose r from them

You need the probability that the 3 selected components at least one are defective.

That is the same as:

(1 - probability that no component of the selection is defective).

The probability that none of the 3 selected components are defective is:

$P = \frac{_{117}C_3}{_{120}C_3}$

Where $_{117}C_3$ is the number of ways to select 3 non-defective components from 117 non-defective components and $_{120}C_3$ is the number of ways to select 3 components from 120.

$_{117}C_3 = 260130$

$_{120}C_3 = 280840$

So:

$P = \frac{260130}{280840} = 0.927$

Finally, the probability that at least one of the selected components is defective is:

$P = 1-0.927 = 0.0737$

P = 7.4%

3 0

3 years ago

The table shows information about the masses of some dogs. a) Work out the minimum number of dogs that could have a mass of more

V125BC [204]

Information regarding various dogs' masses is displayed in the table.

(a) 4 bare minimum number of dogs that could weigh more than 27 kg.

(b) 17 dogs, in total, may weigh more than 27 kg.

Given that,

Information regarding various dogs' masses is displayed in the table.

(a)Calculate the bare minimum number of dogs that could weigh more than 27 kg.

(b)Calculate how many dogs, in total, may weigh more than 27 kg.

(a) 4 bare minimum of dogs that could weigh more than 27 kg.

Because the minimum number so lets assume that all dogs in 20≤x<30 are less then 27.

Therefore, 4 bare minimum of dogs that could weigh more than 27 kg.

(b) 17 dogs, in total, may weigh more than 27 kg.

Because the maximum number so lets assume that all dogs in 20≤x<30 are more then 27.

4+13=17

Therefore,17 dogs, in total, may weigh more than 27 kg.

To learn more about numbers visit: brainly.com/question/17429689

#SPJ1

6 0

1 year ago

According to the weather report, what is the chance of rain or snow?

Viefleur [7K]

Answer:

50? or together will be 90?

5 0

3 years ago

Read 2 more answers

2 factors that add to 6 but multiply to 60

hodyreva [135]

Answer:

The short answer is there isn’t.

Start by writing each of these as an expression:

x * y = 60

x + y = 7

Next, solve each for the same variable; in this case, y:

(x * y) / x = 60 / x

.: y = 60 / x

(x + y) - x = 7 - x

.: y = 7 - x

Next, replace y of the second expression to the first

y = 60 / x & y = 7 - x

.: 7 - x = 60 / x

Now, solve for x:

(7 - x) * x = (60 / x) * x

.: x * 7 - x^2 = 60

This is quadratic, so write it in the form of ax2 + bx + x = 0

(-1)x^2 + (7)x + (-60) = 0

.: a = -1, b = 7, c = -60

Finally solve for b:

x = (-b +- sqrt(b^2 - 4*a*c)) / 2a

.: x = (-7 +- sqrt(7^2 - 4*-1*-60)) / (2 * -1)

.: x = (-7 +- sqrt(49 - 240)) / -2

.: x = (-7 +- sqrt(-191)) / -2

The square root of a negative value is imaginary and thus there’s no real answer to this problem.

8 0

2 years ago

Solve: ( 6 1/5+2 1/2) x 0.85

QveST [7]

It's 21.125 hope this works good luck ;)

6 0

3 years ago

Read 2 more answers