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hammer [34]
3 years ago
12

Repost because i forgot the equation ! 20 points ! someone please help me

Mathematics
2 answers:
Pavlova-9 [17]3 years ago
5 0

Answer:

Infinitely many solutions

Step-by-step explanation:

=> -16x - 20y = 16 => - 4(4x + 5y) = 12

=> 4x + 5y = 12/(-4) => 4x + 5y = -3 .. (1)

=> 8x + 10y = - 6 => 2(4x + 5y) = - 6

=> 4x + 5y = -6/2 => 4x + 5y = - 3 ...(2)

Note that both the equations come out be same on simplification. It means, we are given with only 1 equation that is 4x + 5y = -3, and since there is nothing more about this, equation would give different values of y for infinitely many random values of x.

<u>Thus</u><u>, </u>

<u>E</u><u>quation</u><u> </u><u>has</u><u> </u><u>infinite</u><u>l</u><u>y</u><u> </u><u>many</u><u> </u><u>solution</u><u>s</u><u>.</u>

Dennis_Churaev [7]3 years ago
4 0

Answer:

x=(-3/4)-(5y/4)

Step-by-step explanation:

hello again, the answer is x=(-3/4)-(5y/4). solve using system of equations. please give brainliest

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7 teachers

Step-by-step explanation:

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A recent study reported that high school students spend an average of 94 minutes per day texting. Jenna claims that the average
larisa [96]

Answer:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

Step-by-step explanation:

Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.

From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:

Null Hypothesis: \mu \leq  94

Alternate Hypothesis: \mu > 94

Jenna collected data from a sample of 32 students. So, sample size will be:

Sample Size = n = 32

Sample Mean = x = 96.5

Sample Standard Deviation = s = 6.3

We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.

The formula to calculate the test statistic is:

t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}

Using the values, we get:

t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245

The degrees of freedom will be:

df = n - 1 = 32 - 1 = 31

We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:

p-value = 0.0160

The significance level is:

\alpha =0.05

Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.

Conclusion:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

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Consider the function f(x) = √5x + 15 for the domain [-3, ∞).Find f-¹(x), where f-1 is the inverse of f.Also state the domain of
Ymorist [56]

Answer: We have to find the inverse function of the following:

f(x)=\sqrt{5x+15}\rightarrow[-3,\infty)\Rightarrow(1)

The Inverse of the (1) is as follows:

\begin{gathered} f(x)=y=\sqrt{5x+15} \\  \\ \text{ Switch: }x\text{  and }y\text{ and solve for }y(x)\text{ } \\  \\  \\ y=\sqrt{5x+15}\rightarrow x=\sqrt{5y+15} \\  \\  \\ x^2=5y+15 \\  \\  \\ x^2-15=5y \\  \\  \\ y=\frac{x^2}{5}-\frac{15}{5} \\  \\  \\  \\ y=f^{-1}(x)=\frac{x^2}{5}-3 \\  \\  \\ f^{-1}(x)=\frac{x^{2}}{5}-3 \end{gathered}

The domain of the inverse function is:

x\in(-\infty,+\infty)

3 0
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