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lukranit [14]
3 years ago
11

Let f (x)=X2 -x. if (x)=132, then what is the value of x?

Mathematics
1 answer:
Alex73 [517]3 years ago
7 0
Multiply 132 ..... what does it equal? when an letter is x or another letter it is called a variable. since the variable is next to a number like that with no space you have to multiply the value of the variable and the given number. now finish the problem.
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You have a six-sided die that you roll once. Let Ri denote the event that the roll is i. Let G j denote the event that the roll
Misha Larkins [42]

Answer:

a. P (R3 | G1)=\frac{1}{5}

b. P (R6| G3)= \frac{1}{3}

c. P(G3|E)=\frac{2}{3}

d. P (E|G3)=\frac{2}{3}

Step-by-step explanation:

The sample space associated with the random experiment of throwing a dice is is the equiprobable space {R1, R2, R3, R4, R5, R6}. Then,

a. The conditional probability that 3 is rolled given that the roll is greater than 1? P (R3 | G1) = \frac{P (R3\bigcap G1)}{P(G1)} = \frac{1/6}{5/6} = \frac{1}{5}

b. What is the conditional probability that 6 is rolled given that the roll is greater than 3? P (R6| G3) = \frac{P (R6\bigcap G3)}{P(G3)} = \frac{1/6}{3/6} = \frac{1}{3}

c. What is P [GIE], the conditional probability that the roll is greater than 3 given that the roll is even? P(G3|E) = \frac{P (G3\bigcap E)}{P(E)} = \frac{2/6}{3/6} = \frac{2}{3}

d. Given that the roll is greater than 3, what is the conditional probability that the roll is even? P (E|G3) = \frac{P (E\bigcap G3)}{P(G3)} = \frac{2/6}{3/6} = \frac{2}{3}

6 0
3 years ago
3x3 1/7 which two whole number
pishuonlain [190]

Could you explain a bit more please

7 0
3 years ago
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To save money, a local charity organization wants to target its mailing requests for donations to individuals who are most suppo
mars1129 [50]

Answer:

The 80% confidence interval for difference between two means is (0.85, 1.55).

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for difference between two means is:

CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2,(n_{1}+n_{2}-2)}\times SE_{\bar x_{1}-\bar x_{2}}

Given:

\bar x_{1}=M_{1}=6.1\\\bar x_{2}=M_{2}=4.9\\SE_{\bar x_{1}-\bar x_{2}}=0.25

Confidence level = 80%

t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.20/2, (5+5-2)}=t_{0.10,8}=1.397

*Use a <em>t</em>-table for the critical value.

Compute the 80% confidence interval for difference between two means as follows:

CI=(6.1-4.9)\pm 1.397\times 0.25\\=1.2\pm 0.34925\\=(0.85075, 1.54925)\\\approx(0.85, 1.55)

Thus, the 80% confidence interval for difference between two means is (0.85, 1.55).

3 0
2 years ago
How do i find the value of x on this problem?
zaharov [31]

Answer:

x = 7

Step-by-step explanation:

The proportion is

14/6 = 21/(3x - 12)                 Cross Multiply

14*(3x - 12) = 6*21                 Remove the brackets and combine

42x - 168 = 126                     Add 168 to both sides

42x = 126 + 168

42x =  294                           Divide by 42

x = 294/42

x = 7

8 0
2 years ago
Which equation represents a line which is parallel to x = 0
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I really don’t no so sorry hope u find the asset if you haven’t
7 0
3 years ago
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