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bearhunter [10]
3 years ago
14

To earn an A in an algebra course, a student must have a test average of at least 90. Mary has grades of 95, 82, 88 on her first

three algebra tests. What minimum score does Mary need to make on her fourth test to earn an A in her algebra course?
Mathematics
1 answer:
Fofino [41]3 years ago
5 0

Answer: Mary need to make at-least 95 on her fourth test to earn an A in her algebra course.

Step-by-step explanation:

Let x be the grades scored by Mary in the fourth algebra test.

Mary has grades of 95, 82, 88 on her first three algebra tests.

Then, the combined scores in four test will become = 95+82+88+x =  265+x

Average score = (Sum of all scores) ÷ (Number of tests)

=\dfrac{265+x}{4}

As per given ,

To earn an A in an algebra course, a student must have a test average of at least 90.

i.e. Average score ≥ 90

\Rightarrow\ \dfrac{265+x}{4}\geq90\\\\\Rightarrow\ 265+x\geq 90\times4=360\\\\\Rightarrow\ x\geq360-265 =95\\\\\Rightarrow\ x\geq90

Hence, Mary need to make at-least 95 on her fourth test to earn an A in her algebra course.

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M=2,000÷((1−(1+0.072÷12)^(−12
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8 0
3 years ago
Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are
Svetlanka [38]

Answer:

(a) Number of sets B given that

  • A⊆B⊆C: 2¹⁰.  (That is: A is a subset of B, B is a subset of C. B might be equal to C)
  • A⊂B⊂C: 2¹⁰ - 2.  (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

<h3>(a)</h3>

Let x_1, x_2, \cdots, x_{20} denote the 20 elements of set X.

Let x_1, x_2, \cdots, x_{10} denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain x_1, x_2, \cdots, x_{10}.

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from x_1, x_2, \cdots, x_{20}.

\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}&  \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}.

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus 2^{10} = 1024 possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves 2^{10} -2 = 1024 - 2 = 1022 possibilities.

<h3>(b)</h3>

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}&  \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}.

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus 2^{10} = 1024 possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's 2^{20} = 1048576 possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only 2^{20} - 2^{10} possibilities left.

5 0
3 years ago
An investment website can tell what devices are used to access their site. The site managers wonder whether they should enhance
lbvjy [14]

Answer: a)  Unimodal and symmetric

b) 0.26

c) 0.038

Step-by-step explanation:

Given: Sample size of investors (n)= 131

True proportion of smartphone users(p)  =26%

a) Since sampling distribution for the sample proportion is approximately normal when n is larger.

Normal distribution is Unimodal and symmetric.

So correct option : Unimodal and symmetric

b)  mean of this sampling​ distribution = p = 0.26

c)  standard deviation of the sampling​distribution = \sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.26\times (1-0.26)}{131}}=\sqrt{0.00146870229008}

=0.0383236518364\approx0.038

6 0
3 years ago
(5÷3)x+(1÷2)=0 <br><br>find x​
xxMikexx [17]
(1.6666667)x+(.5)=0
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x= -.5/1.6666667
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8 0
3 years ago
What is the probability that a randomly selected person is on the swim team? Express the probability as a percent rounded to the
frez [133]
<span>What is the probability that a randomly selected person is on the swim team?
28 out of 67 people are on the swim team
so that's 28/67
</span><span>41.7910448% round to the nearest tenth 
41.8% is the answer. 

Hope this helps, Have a nice day!</span>
5 0
3 years ago
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