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Reika [66]
3 years ago
5

What property is illustrated in the equation? 7 + 1/2 = 1/2 + 7

Mathematics
1 answer:
aliya0001 [1]3 years ago
3 0

Communicative

Hope this helps!

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Given: f(x) = X^2 - 3 and g(x) = X + 1 The composite function g°f is:_______.
LenKa [72]

Answer:

x² - 2

Step-by-step explanation:

To obtain (g ○ f)(x), substitute x = f(x) into g(x), that is

g(x² - 3)

= x² - 3 + 1

= x² - 2

8 0
3 years ago
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Help me please what does y equal ?
kramer

Answer:

y= 5

Step-by-step explanation:

:)

8 0
3 years ago
What two numbers multiply by 25 but adds up to 6
Natalka [10]
No two <em>intergers</em> can solve this problem. If you're not in Algebra II yet, the answer is probably "none."

Just look at the factor pairs of 25:
1 × 25
-1 × 25
5 × 5
-5 × -5
Clearly none of those can add up to 6.

If you want the more complex answer, I'll show you how here. If you don't understand why it doesn't work, that's okay. I just want you to see that  there's not an actual answer to the problem.

a+b = 6
a = 6-b
ab = 25
(6-b)b = 25
6b -b² = 25
-b² + 6b = 25
b² -6b = -25
Factor by splitting the middle.
Half of -6 is -3, (-3)² = 9. Add this to each side.
b² -6b + 9 = -16
Factor the perfect square trinomial.
(b-3)² = -16
Take the square root of each side.
b-3 = 4i
b = 3+4i
a+b = 6
a+3+4i = 6
a= 3-4i
<em>(The "i" stands for an imaginary number, specifically, the square root of -1.)</em>
8 0
3 years ago
Many elementary school students in a school district currently have ear infections. A random sample of children in two different
Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school n_1 = 42

sample size for the second school n_2  = 34

so 16 out of 42 i.e x_1 = 16 and 18 out of 34 i.e x_2 = 18 have ear infection.

the proportion of students with ear infection Is as follows:

\hat p_1 = \dfrac{16}{42} = 0.38095

\hat p_2 = \dfrac{18}{34}  =  0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}

\bar p = \dfrac{16 +18}{42 +34}

\bar p = \dfrac{34}{76}

\bar p = 0.447368

\bar p + \bar  q = 1 \\ \\ \bar q = 1 -\bar  p \\  \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632

The pooled standard error can be computed by using the formula:

S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} +  \dfrac{\bar p \bar p}{n_2} }

S.E = \sqrt{ \dfrac{  0.447368 *  0.552632}{ 42} +  \dfrac{ 0.447368 *  0.447368}{34} }

S.E = \sqrt{ \dfrac{  0.2472298726}{ 42} +  \dfrac{ 0.2001381274}{34} }

S.E = \sqrt{ 0.01177284105}

S.E = 0.1085

The test statistics is ;

z = \dfrac{\hat p_1 - \hat p_2}{S.E}

z = \dfrac{0.38095- 0.5294}{0.1085}

z = \dfrac{-0.14845}{0.1085}

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

5 0
3 years ago
Help plzzzzzzzzzzzzzz ​
a_sh-v [17]

Answer:

ok

Step-by-step explanation:

8 0
3 years ago
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