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GREYUIT [131]
3 years ago
8

At the calendar shop, wall calendars cost $8 and desk calendars cost $10. Steve spent $42 to buy 5 calendars. How many of each t

ype of calendar did Steve buy?
______ Wall Calendar

______ Desk Calender
Mathematics
2 answers:
alexira [117]3 years ago
6 0
4 wall calendars and 1 desk calendar
kodGreya [7K]3 years ago
3 0
He bought 4 wall calendars and 1 desk calendars

Because 4 X 8 = 32
And 1 x 10 = 10

32 + 10 = 42
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Evaluate the determinant for the following matrix: 1 4 4 5 2 2 1 5 5
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A Laplace expansion along the first row gives

\begin{vmatrix}1&4&4\\5&2&1\\1&5&5\end{vmatrix}=1\begin{vmatrix}2&1\\5&5\end{vmatrix}-4\begin{vmatrix}5&1\\1&5\end{vmatrix}+4\begin{vmatrix}5&2\\1&5\end{vmatrix}
=1(10-5)-4(25-1)+4(25-2)
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3 years ago
coach cliffman made a deposit of $1,800 into an account that earns 2% annual simple interest. find the amount of interest that c
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He earns $36 after 3 years
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2 years ago
What is 3x1+1+1+1+1+1+1+1+2+3
Andrej [43]

Answer:

15

Step-by-step explanation:

Math is fun.

7 0
3 years ago
Read 2 more answers
1. The national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15. The dean of a college wants to kno
Nat2105 [25]

Answer:

We conclude that the mean IQ of her students is different from the national average.

Step-by-step explanation:

We are given that the national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15.

The dean of a college want to test whether the mean IQ of her students is different from the national average. For this, she administers IQ tests to her 144 students and calculates a mean score of 113

Let, Null Hypothesis, H_0 : \mu = 100 {means that the mean IQ of her students is same as of national average}

Alternate Hypothesis, H_1 : \mu\neq 100  {means that the mean IQ of her students is different from the national average}

(a) The test statistics that will be used here is One sample z-test statistics;

               T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~ N(0,1)

where, Xbar = sample mean score = 113

              s = population standard deviation = 15

             n = sample of students = 144

So, test statistics = \frac{113-100}{\frac{15}{\sqrt{144} } }

                             = 10.4

Now, at 0.05 significance level, the z table gives critical value of 1.96. Since our test statistics is more than the critical value of z which means our test statistics will fall in the rejection region and we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean IQ of her students is different from the national average.

3 0
3 years ago
Answer to this please ?
stepan [7]

Answer:

(-a,b)

Step-by-step explanation:

You use midpoint formula which is (x1+x2/2),(y1+y2/2)

8 0
3 years ago
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