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3 years ago

Y=x-4 y=4x+2 whats the solution of system equations

Mathematics

1 answer:

Fittoniya [83]3 years ago

6 0

Answer:

X=-2

Y=-6

Step-by-step explanation:

Since both equations are already expressed in the simplest form of y, then we equate them to be equal hence

x-4=4x+2

Bringing like terms together

-4-2=4x-x

Solving both sides

-6=3x

Making x the subject then

X=-6/3=-2

Subsrituting the value of x into any of the initial equations

Y=x-4 then y=-2-4=-6

Therefore, the solution is

X=-2

Y=-6

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Answer:

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When Nabhitha goes bowling, her scores are normally distributed with a mean of 115

frutty [35]

Answer:

By the Empirical Rule, in 99.7% of the games that Aubree bowls she scores between 148 and 232Step-by-step explanation:The Empirical Rule states that, for a normally distributed random variable:68% of the measures are within 1 standard deviation of the mean.95% of the measures are within 2 standard deviation of the mean.99.7% of the measures are within 3 standard deviations of the mean.In this problem, we have that:Mean = 190Standard deviation = 14Using the empirical rule, what percentage of the games that Aubree bowls does she score between 148 and 232?148 = 190 - 3*14So 148 is 3 standard deviations below the mean.232 = 190 + 3*14So 232 is 3 standard deviations above the meanBy the Empirical Rule, in 99.7% of the games that Aubree bowls she scores between 148 and 232

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2 years ago

Mendel crossed tall pea plants with short pea plants. The offspring were all tall plants.

melomori [17]

Answer:

Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

A box of Munchkins contains chocolate and glazed donut holes. If Jacob ate 2 chocolate

neonofarm [45]

Answer:

24 munchkins.

Step-by-step explanation:

Let C be the number of chocolate and D be number of glazed donut holes in the original box.

We are told if Jacob ate 2 chocolate munchkins, then 1/11 of the remaining Munchkins would be chocolate. We can represent this information as:

$C-2=\frac{1}{11}*(C+D-2)...(1)$

We are also told if he instead added 4 glazed Munchkins to the original box, 1/7 of the Munchkins would be chocolate. We can represent this information as:

$C=\frac{1}{7}*(C+D+4)...(2)$

Upon substituting C's value from equation (2) in equation (1) we will get,

$\frac{1}{7}*(C+D+4)-2=\frac{1}{11}*(C+D-2)$

Let us have a common denominator on right side of equation.

$\frac{1}{7}*(C+D+4)-\frac{7*2}{7}=\frac{1}{11}*(C+D-2)$

$\frac{C+D+4-14}{7}=\frac{1}{11}*(C+D-2)$

Multiplying both sides of our equation by 7, we will get,

$7*\frac{C+D-10}{7}=7*\frac{1}{11}*(C+D-2)$

$C+D-10=\frac{7}{11}*(C+D-2)$

Multiplying both sides of our equation by 11, we will get,

$11*(C+D-10)=11*\frac{7}{11}*(C+D-2)$

$11*(C+D-10)=7*(C+D-2)$

$11C+11D-110=7C+7D-14$

$11C-7C+11D-7D=-14+110$

$4C+4D=96$

$4(C+D)=96$

$(C+D)=\frac{96}{4}$

$(C+D)=24$

Therefore, the total number of Munchkins in original box is 24.

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