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spin [16.1K]
3 years ago
5

Your history test is out of 30 points. How many points would you need to get if you wanted a ‘B’ or higher? (Hint: A grade of 80

% would be a ‘B’.) *
Mathematics
2 answers:
Sloan [31]3 years ago
7 0
To get a B, you need at least 80%.

30 x 0.8 = 24

80% of 30 is 24, so you would need to get at least 24 points <3
Alika [10]3 years ago
4 0
If you want to get a B you would have to get an 80%.

30 x 0.8 (80%) = 24
You would need to get 24 points
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Multiplying a trinomial by a trinomial follows the same steps as multiplying a binomial by a trinomial. Determine the degree and
FinnZ [79.3K]

Answer: Degree of polynomial (highest degree) =4

Maximum possible terms =9

Number of terms in the product = 5

Step-by-step explanation:

A trinomial is a polynomial with 3 terms.

The given product of trinomial: (x^2 + x + 2)(x^2 - 2x + 3)

By using distributive property: a(b+c+d)= ab+ac+ad

(x^2 + x + 2)(x^2 - 2x + 3)=(x^2 + x + 2) x^2+(x^2 + x + 2) (-2x)+(x^2 + x + 2)(3)\\\\=x^2(x^2)+x(x^2)+2(x^2)+x^2 (-2x)+x (-2x)+2 (-2x)+x^2 (3)+x (3)+2 (3)\\\\\\=x^4+x^3+2x^2-2x^3-2x^2-4x+3x^2+3x+6

Maximum possible terms =9

Combine like terms

x^4+x^3-2x^3+3x^2-4x+3x+6\\\\=x^4-x^3+3x^2-x+6

Hence, \left(x^2\:+\:x\:+\:2\right)\left(x^2\:-\:2x\:+\:3\right)=x^4-x^3+3x^2-x+6

Degree of polynomial (highest degree) =4

Number of terms = 5

6 0
2 years ago
A bakery sells 6350 muffins in 2010. The bakery sells 8310 muffins in 2015. Write a linear model that represents the number y of
krok68 [10]

Answer:

y = 6,350 + 392x

Step-by-step explanation:

3 0
3 years ago
A negative number divided by a positive number is:
Verdich [7]

Answer:

When you divide a negative number by a positive number then the quotient is negative.

Step-by-step explanation:

hope this helped

7 0
2 years ago
Hal has a square garden in his backyard with an area of 210 square feet. To the nearest half foot, what are the dimensions of th
Zanzabum

I believe the answer is 57sqft

7 0
3 years ago
Confidence Interval Mistakes and Misunderstandings—Suppose that 500 randomly selected recent graduates of a university were as
kvv77 [185]

Answer:

The correct 95% confidence interval is (8.4, 8.8).

Step-by-step explanation:

The information provided is:

n=500\\\bar x=8.6\\\sigma=2.2

(a)

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The 95% confidence interval for the average satisfaction score is computed as:

8.6 ± 1.96 (2.2)

This confidence interval is incorrect.

Because the critical value is multiplied directly by the standard deviation.

The correct interval is:

8.6\pm 1.96 (\frac{2.2}{\sqrt{500}})=8.6\pm 0.20=(8.4,\ 8.6)

(b)

The (1 - <em>α</em>)% confidence interval for the parameter implies that there is (1 - <em>α</em>)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.

The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.

(c)

The (1 - <em>α</em>)% confidence interval for population parameter implies that there is a (1 - <em>α</em>) probability that the true value of the parameter is included in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.

Thus, the students is not making any misinterpretation.

(d)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

In this case the sample size is,

<em>n </em>= 500 > 30

Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.

7 0
3 years ago
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