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3 years ago

Can pretty please help me with 2 math question its due today

Mathematics

2 answers:

Mashcka [7]3 years ago

5 0

Answer:
(4x + 12 ) - ( -8 ) =
4x + 20
Explanation:
Just add them up and form a equation ( this is for the second one )

Yakvenalex [24]3 years ago

5 0

(4x + 12) - (-8)=
hope this helps

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Black_prince [1.1K]

Answer:

x = 50

Step-by-step explanation:

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3 years ago

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A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

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3 years ago

Which equation represents the polynomial function with zeros −1, 1, and 3 (multiplicity of 2), and a y-intercept of −18? (4 poin

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Correct Answer:
3rd option is the correct answer

Solution:
The zeros of the polynomial are -1,1 and 3. The multiplicity of 3 is 2. So the polynomial can be expressed as:

$y=a (x-3)^{2}(x-1)(x+1)$

The y-intercept of the polynomial is -18. This means the polynomial passes through the point (0,-18). Therefore, y must be -18 when x = 0. Using these values of x and y in previous equation we get:

$-18=a (0-3)^{2}(0-1)(0+1) \\ \\ 
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The final equation of the polynomial becomes:

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3 years ago

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