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OleMash [197]
3 years ago
5

A high school basketball game between the Raiders and the Wildcats was tied at the end of the first quarter. The number of point

s scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than 100 points. What was the total number of points scored by the two teams in the first half
Mathematics
1 answer:
AnnyKZ [126]3 years ago
4 0

Answer: 5 and 14.

Step-by-step explanation:

We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let a,a+d,a+2d,a+3d be the quarterly scores for the Wildcats. The sum of the Raiders scores is a(1+r+r^{2}+r^{3}) and the sum of the Wildcats scores is 4a+6d. Now we can narrow our search for the values of a,d, and r. Because points are always measured in positive integers, we can conclude that a and d are positive integers. We can also conclude that $r$ is a positive integer by writing down the equation:

a(1+r+r^{2}+r^{3})=4a+6d+1

Now we can start trying out some values of r. We try r=2, which gives

15a=4a+6d+1

11a=6d+1

We need the smallest multiple of 11 (to satisfy the <100 condition) that is 1 (mod 6). We see that this is 55, and therefore a=5 and d=9.

So the Raiders' first two scores were 5 and 10 and the Wildcats' first two scores were 5 and 14.

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AnnZ [28]
The answer of 8+11=40
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3 years ago
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Prove the identity 2csc2x=csc^2xtanz
Verizon [17]

Step-by-step explanation:

Consider the LHS, after the 5th step, consider the RHS

2 \csc(2x)  =  \csc {}^{2} (x)  \tan(x)

2 \frac{1}{ \sin(2x) }  =  \csc {}^{2} (x)  \tan(x)

2  \times \frac{1}{2 \sin(x)  \cos(x) }  =  \csc {}^{2} (x)  \tan(x)

\frac{1}{ \sin(x) \cos(x)  }  =   \csc {}^{2} (x)  \tan(x)

\csc(x)  \sec(x)  =  \csc {}^{2} (x)  \tan(x)

Consider the RHS

\csc(x)  \sec(x)  =( 1 +  \cot {}^{2} (x) ( \tan(x))

\csc(x)  \sec(x)  =  \tan(x)  +  \cot(x)

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7 0
2 years ago
Hi guys i cant get this question can yous please help me its due in two days
Natasha2012 [34]

Answer:

(2400+420).

Step-by-step explanation:

We can break this into the same pattern as the other numbers, giving us:

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When we multiply these we get:

(2400+420).

7 0
4 years ago
Given the polynomial 2x^3+18x^2-18x-162, what is the value of the constant "k" in the factored form?
maxonik [38]
2x³+18x²<span>-18x-162 = 
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7 0
3 years ago
Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar
yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.  

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

<em>Let </em>\mu_1<em> = true mean number of sales at location A.</em>

<em />\mu_2 = <em>true mean number of sales at location B</em>

So, Null Hypothesis, H_0 : \mu_1-\mu_2\geq0  or  \mu_1 \geq \mu_2     {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, H_A : \mu_1-\mu_2  or  \mu_1< \mu_2    {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

                        T.S. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t_n__1_-_n__2-2

where, \bar X_1 = sample average of items sold at location A = 39

\bar X_2 = sample average of items sold at location B = 49

s_1 = sample standard deviation of items sold at location A = 8

s_2 = sample standard deviation of items sold at location B = 4

n_1 = sample of days location A was observed = 18

n_2 = sample of days location B was observed = 13

Also,  s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} }  = \sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2}  }{18+13-2} }  = 6.64

So, <u><em>test statistics</em></u>  =  \frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13}  } }  ~ t_2_9  

                               =  -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

<em>Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

3 0
3 years ago
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