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motikmotik
4 years ago
7

Can any one help me and explain please

Mathematics
1 answer:
dangina [55]4 years ago
8 0
(2x - 5)( 4 {x}^{2} - 3x + 1) = \\ = 8 {x}^{3} - 6 {x}^{2} + 2x - 20 {x}^{2} + 15x - 5 = \\ = 8 {x}^{3} - 26 {x}^{2} + 17x - 5
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HELP FAST!!! Here are two rectangles. The length and width of one rectangle are 8 and 5. The width of the other rectangle is 5,
deff fn [24]

Answer:

5•x=? and 8•5=40, personal opinion: the left rectangle is most likely a square so if you need to solve it it's 5•5=10

Step-by-step explanation:

A(area)=l (length)•w (width)

just plug in your numbers, in this case it would be 8(length) times 5(width)

6 0
3 years ago
(This is literally 7th grade math dude should be easy for most of you)
Korolek [52]

Answer:

a. no b. 40.05

Step-by-step explanation:

a. to figure out the number of goals per game you have to divide the total amount of goals by the number of games:

last season: 41 ÷ 15 = 2.73

this season: 24 ÷ 9 = 2.67

as you can see the answer to the last season equation is higher than the this season meaning the player scored more goals last season.

b. we already know that the amount of games the player won each game (all of them supposedly being exactly even) is 2.67. so what we have to do is multiply it by 15 the supposed number of games the player is playing this season to get the total number of goals they will get for the entire season:

2.67 x 15 = 40.05 = the prediction of the total goals of the player

7 0
3 years ago
Read 2 more answers
HEY CAN ANYONE PLS ANSWER DIS MATH QUESTION!!!!!!!!
ziro4ka [17]

Answer:

  • 80 tons
  • $20,000

Step-by-step explanation:

<u>21. 1st scenario</u>

  • B = fy + wz
  • 100,000 = 15,000*4 + 500w
  • 500w = 100,000 - 60,000
  • 500w = 40,000
  • w = 40,000/500
  • w = 80

80 tons of wheat

<u>22. 2nd scenario</u>

  • B = fy + wz
  • 50,000 = 2y + 200*50
  • 50,000 = 2y + 10,000
  • 2y = 50,000 - 10,000
  • 2y = 40,000
  • y = 20,000

Fighter plane's price is $20,000

8 0
3 years ago
A theory predicts that the mean age of stars within a particular type of star cluster is 3.3 billion years, with a standard devi
Luden [163]

Answer:

Null hypothesis:\mu \leq 3.3  

Alternative hypothesis:\mu > 3.3  

z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768  

Since is a one right tailed test the p value would be:  

p_v =P(z>1.768)=0.039  

Step-by-step explanation:

Data given and notation

\bar X=3.4 represent the sample mean  

\sigma=0.4 represent the population deviation for the sample

n=50 sample size  

\mu_o =3.3 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 3.3, the system of hypothesis would be:  

Null hypothesis:\mu \leq 3.3  

Alternative hypothesis:\mu > 3.3  

Compute the test statistic  

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768  

P value

Since is a one right tailed test the p value would be:  

p_v =P(z>1.768)=0.039  

6 0
3 years ago
*brainliest for best answer thank you* Solve for b.
vekshin1

Hello from MrBillDoesMath!

Answer:

The fourth choice,  b = +\- sqrt( sg + a^2)

Discussion:

s = (b^2 - a^2)/g     =>              multiply both sides by "g"

sg = b^2 - a^2        =>               add a^2 to both sides

sg + a^2 = b^2       =>                take the square root of each side

b = +\- sqrt( sg + a^2)


which is the fourth choice.


Thank you,

MrB

7 0
3 years ago
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