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egoroff_w [7]
3 years ago
8

Which Set of Ordered Pairs Doesn’t Represent A Function DUE TONIGHT WILL GIVE BRAINLIEST

Mathematics
1 answer:
anygoal [31]3 years ago
4 0

Answer:

The sewer is

Step-by-step explanation:

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Solve the equation 3x - 7y equals -28 for y
SashulF [63]

thats wrong the answer is 35y/3


6 0
3 years ago
Work out the shaded area​
Dennis_Churaev [7]

Answer:

<h2> 70 {cm}^{2}</h2>

Step-by-step explanation:

Shaded area:

Area of big rectangle - Area of small rectangle

= 10 \times 12 - (10 - 5) \times (12 - 2)

Calculate the difference

= 10  \times 12 - 5 \times 10

Calculate the product

= 120 - 50

Calculate the difference

= 70 \:  {cm}^{2}

Hope this helps...

Best regards!!

7 0
3 years ago
Ray wants to buy the larger of two aquariums. one aquarium has a base that is 20 inches by 20 inches and height that is 18 inche
Svetradugi [14.3K]
Ray should buy the first one because it has the greater volume by 1,440 square inches.
Tank 1: 20x20x18=7,200
Tank 2: 40x12x12=5,760
7,200-5,760=1,440
4 0
3 years ago
Read 2 more answers
How many ways can you select a committee of 4 students out of 10 students?
Aleks04 [339]

\binom{10}{4}=\frac{10!}{6!4!}=\frac{10(9)(8)(7)}{24}=10(3)(7)=\boxed{210}

3 0
1 year ago
John and Martha are contemplating having children, but John’s brother has galactosemia (an autosomal recessive disease) and Mart
Rina8888 [55]
<h2>Answer:</h2>

Probability=\frac{1}{24}

<h2>Step-by-step explanation:</h2>

As the question states,

John's brother has Galactosemia which states that his parents were both the carriers.

Therefore, the chances for the John to have the disease is = 2/3

Now,

Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.

Now, one of those children married with a person.

So,

Probability for the child to have disease will be = 1/2

Now, again the child's child (Martha) probability for having the disease is = 1/2.

Therefore,

<u>The total probability for Martha's first child to be diagnosed with Galactosemia will be,</u>

Probability=\frac{2}{3}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{4}\\Probability=\frac{1}{24}

(Here, we assumed that the child has the disease therefore, the probability was taken to be = 1/4.)

<em><u>Hence, the probability for the first child to have Galactosemia is \frac{1}{24}</u></em>

3 0
3 years ago
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