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VLD [36.1K]
2 years ago
14

Which equation is equivalent to StartRoot x EndRoot + 11 = 15?

Mathematics
1 answer:
Lerok [7]2 years ago
5 0

The equivalent expression of \sqrt x + 11 = 15 is \sqrt x = 15 - 11

<h3>How to determine the equivalent expression?</h3>

The complete question is in the attached image

The expression is given as:

\sqrt x + 11 = 15

Subtract 11 from both sides of the equation

\sqrt x + 11 - 11= 15 - 11

This gives

\sqrt x = 15 - 11

Hence, the equivalent expression of \sqrt x + 11 = 15 is \sqrt x = 15 - 11

Read more about equivalent expression at:

brainly.com/question/4344214

#SPJ1

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monitta
Because every x value cannot have more than one y value so doing it will make sure it has only one value on the line.
7 0
4 years ago
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Choose the three equivalent expressions: (Select all that apply 6( 4x + 5y) 20x + 30y 2(10x + 3y) 10(2x) + 10( 3y) 10(2x + 3y)​
mart [117]

Answer

Option 2, 4 and 5 :)

Step-by-step explanation:

Option 2 is already simplified to 20x + 30y

Option 4,

10*2x=20x and 10*3y= 30y so option 4 also simplifies to 20x + 30y

Next Option 5, 10*2x= 20x and 10*3y=30y. So it's also 20x + 30y

4 0
3 years ago
Is the mean weight of female college students still 59.4 kg? To test this, you take a random sample of 20 students, finding a me
adell [148]

Answer:

Correct option: (B)

Step-by-step explanation:

A one-sample <em>t</em>-test can be performed to determine whether the mean weight of female college students is still 59.4 kg.

The hypothesis can be defined as:

<em>H₀</em>: The mean weight of female college students has changed.

<em>Hₐ</em>: The mean weight of female college students has not changed.

The information provided is:

\bar x=61.65\\s=6.39\\n=20\\\alpha =0.10

As the there is no information about the population standard deviation we will use a <em>t</em>-test for the mean.

The test statistic is:

t=\frac{\bar x-\mu}{\s/\sqrt{n}}=\frac{61.65-59.4}{6.39/\sqrt{20}}=1.57

Decision rule:

If the if the <em>p</em>-value of the test is less than the significance level of the test <em>α</em> then the null hypothesis will be rejected and vice versa.

The degrees of freedom of the test is:

df=n-1=20-1=19

The test is two-tailed.

Compute the <em>p</em>-value of the test as follows:

p-value=2\times P(t_{0.10/2, 19}

*Use a <em>t</em>-table for the probability.

The <em>p</em>-value = 0.13 > <em>α</em> = 0.10

The null hypothesis was failed to be rejected.

As the null hypothesis was rejected it can be concluded that there is not sufficient evidence that the mean weight of female students has changed.

6 0
4 years ago
[THESE ARE NOT THE RIGHT ANSWERS I JUST GUESSED]
marishachu [46]

The parent function of the function represented in the table is <u>exponential</u>.

  • Note if you subtract 10 from each output, you get f(x)=3^x.

If function f was translated down 4 units, the <u>f(x)</u> values would be <u>decreased by 4</u>.

  • See attached image.

A point on the table for the transformed function would be <u>(4, 87).</u>

  • f(4)-4=87

5 0
2 years ago
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devlian [24]

Answer:

Gracias

Step-by-step explanation:

7 0
3 years ago
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