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Oksanka [162]
2 years ago
11

Which expression is equivalent to 8xy + xy + 19xy+X?

Mathematics
1 answer:
Viktor [21]2 years ago
4 0

Answer:

0

Step-by-step explanation:

no mo explanation for this next time

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A scientist testing the effects of a chemical on apple yield (apples/acre) sprays an orchard with the chemical. A second orchard
JulsSmile [24]

Following are the dependent variables:

<em>1. The amount of water that each orchard receives.</em>

<em>2. The species of trees in the orchard.</em>

Reason:

The exercise scientist is looking for the effects of a chemical between an apple crop to which it is administered and another to which it is not, 4 options are presented, of which it is essential to count as a variable the amount of water each Orchard and tree species in the orchard, since they can generate alterations in the results, the other two variables of the exercise such as number of apples and size of the orchards are not significant and their variations do not affect the scientist's objective.

Learn more about Dependent Variable on:

brainly.com/question/1670595

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2 years ago
Can someone please help me
Lilit [14]

Answer:

9.2

Step-by-step explanation:

Use law of sines

a/sinA = b/sinB

5/sin 31 = b/sin 108

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A countertop is 16 feet long and 3 feet wide. what is the area of the countertop in square meters
Paladinen [302]

The area of the countertop is (16-ft x 3-ft) = 48 feet² .

1 foot = 0.3048 meter (rounded)

(1 foot)² = (0.3048 meter)² = 0.0929 meter²

48 feet² = (48 x 0.0929) =  4.459 meter² (rounded)

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frez [133]

Answer:

A B and C are true

Step-by-step explanation:

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A shipment of 11 printers contains 2 that are defective. Find the probability that a sample of size 2​, drawn from the 11​, will
svet-max [94.6K]

The required probability is \frac{36}{55}

<u>Solution:</u>

Given, a shipment of 11 printers contains 2 that are defective.  

We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.

Now, we know that, \text { probability }=\frac{\text { favourable outcomes }}{\text { total outcomes }}

Probability for first draw to be non-defective =\frac{11-2}{11}=\frac{9}{11}

(total printers = 11; total defective printers = 2)

Probability for second draw to be non defective =\frac{10-2}{10}=\frac{8}{10}=\frac{4}{5}

(printers after first slot = 10; total defective printers = 2)

Then, total probability =\frac{9}{11} \times \frac{4}{5}=\frac{36}{55}

7 0
3 years ago
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