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damaskus [11]
3 years ago
12

Match each number on the left with the correct description of the number on the right.

Mathematics
1 answer:
7nadin3 [17]3 years ago
5 0

Answer:

I hope this helps you

:)

You might be interested in
)
Leto [7]

Answer:

Number of Adult's tickets sold on Saturday = 3,356

Number of Children's tickets sold on Saturday = 2, 928

Total number of tickets sold over these two days is 8,938.

Step-by-step explanation:

Here, the number of tickets sold on FRIDAY:

Adult Ticket sold = 1,678

Children's Tickets sold = 976

So, the total number of tickets sold on Friday  

= Sum of ( Adult + Children's ) tickets  = 1,678  + 976 = 2,654 ....  (1)

The number of tickets sold on SATURDAY:

Adult Ticket sold =   2 times  the number of adult tickets sold on Friday

                            =  1,678 x 2  = 3,356

Children's Tickets sold = 3 times the number of children's tickets sold on Friday.

=  976  x 3 = 2, 928

So, the total number of tickets sold on Saturday  

= Sum of ( Adult + Children's ) tickets  = 3,356 + 2,928 = 6, 284 ....  (2)

Now, the total number of tickets booked in these two days :

Sum of tickets booked on (Friday + Saturday)

= 2,654 +  6, 284  =   8,938

Hence, total number of tickets sold over these two days is 8,938.

6 0
3 years ago
WIL GIVE BRAINLIEST<br> Focus (- 2, - 3) Directrix x = 6
aliina [53]

Answer:

Here's ya answer!

(

y

+

3

)

2

=

−

6

(

x

−

7

2

)

Sorry its upside down, but good luck on that test!

4 0
3 years ago
Find the product of (x − 3)2.<br> x2 + 6x + 9<br> x2 − 9<br> x2 − 6x + 9<br> x2 + 9
Simora [160]
(x-3)² ⇒ (x-3)(x-3)

(x-3)(x-3) = x(x-3) -3(x-3) ⇒ x² - 3x - 3x + 9 ⇒ x² - 6x + 9

The 3rd option is the correct answer. x
² - 6x + 9
3 0
4 years ago
Read 2 more answers
Given that p²,q² are two roots of x²-x+16=0. <br>Form another equation with roots 1/p,1/q.
statuscvo [17]

Answer:

  x² -3/4x +1/4 = 0

Step-by-step explanation:

Consider the two equations in factored and expanded forms:

  (x -p²)(x -q²) = x² -(p²+q²)x +p²q² = 0   ⇒   p²+q² = 1, p²q² = 16

and

  (x -1/p)(x -1/q) = x² -(1/p+1/q)x +1/(pq) = 0

Consider the squares of the sum and product of roots:

  constant term: (1/(pq))² = 1/(p²q²) = 1/16   ⇒   1/(pq) = √(1/16) = 1/4

  x-term: (1/p +1/q)² = (p +q)²/(pq)² = (p² +q² +2pq)/(p²q²)

  = (p² +q²)/(p²q²) +2/(pq)

  = 1/16 +2/√16 = 9/16   ⇒   (1/p +1/q) = √(9/16) = 3/4

Then the equation with roots 1/p and 1/q is ...

  x² -3/4x +1/4 = 0

6 0
3 years ago
The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for t
nadezda [96]

Answer:

a. There is evidence that the population mean is above $300.

b. There is no evidence that the population mean is above $300.

c. There is no evidence that the population mean is above $300.

d. The director could ask for cheaper similar books.

Step-by-step explanation:

Let X be the random variable that represents the cost of textbooks. We have observed n = 25 values, \bar{x} = 315.4 and s = 43.20. We suppose that X is normally distributed.

We have the following null and alternative hypothesis

H_{0}: \mu = 300 vs H_{1}: \mu > 300 (upper-tail alternative)

We will use the test statistic

T = \frac{\bar{X}-300}{S/\sqrt{25}} and the observed value is

t_{0} = \frac{315.4 - 300}{43.20/\sqrt{25}} = 1.7824.

If H_{0} is true, then T has a t distribution with n-1 = 24 degrees of freedom.

a. The rejection region is given by RR = {t | t > t_{0.9}} where t_{0.9} = 1.3178 is the 90th quantile of the t distribution with 24 df, so, RR = {t | t > 1.3178}. Because the observed value satisty 1.7824 > 1.3178, there is evidence that the population mean is above $300.

b. If s = 75, then the observed value is t_{0} = \frac{315.4 - 300}{75/\sqrt{25}} = 1.0267. The rejection region for a 0.05 level of significance is RR = {t | t > t_{0.95}} where t_{0.95} = 1.7108 is the 95th quantile of the t distribution with 24 df, so, RR = {t | t > 1.7108}. Because the observed value does not fall inside the rejection region, there is no evidence that the population mean is above $300.

c. If \bar{x} = 305.11 and s = 43.20, the observed value is t_{0} = \frac{305.11 - 300}{43.20/\sqrt{25}} =  0.5914. For RR = {t | t > 1.3178} we have that the observed value does not fall inside RR, therefore, there is no evidence that the population mean is above $300.

d. Because the director of admissions is concerned about the high cost of textbooks, and there is evidence that the population mean of costs is above $300, the director could ask for cheaper similar books.

8 0
3 years ago
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