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butalik [34]
3 years ago
12

Please answer very urgent!! answer using a / only 1 number for each slot example:3/4: wrong tho just random number but yeah plea

se put the correct answer

Mathematics
1 answer:
Naya [18.7K]3 years ago
6 0

Answer:

5/6

Step-by-step explanation:

to subtract fractions you change both fractions to the same denominator. The website arealdy did it for you. The second step is to subtract the top and put it over the common denominator. 8-3 is 5. 5/6

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Show me how to multiply .54 times .5
saul85 [17]
First do 54x5 which is equal to 270. then, you move the decimal point to the left 3 times; 2 for the .54 and 1 for the .5. you will then get 0.27 which is ur answer
6 0
3 years ago
What would these fractions be in percent form? <br> 1.30/20 <br> 10.26/20
Eduardwww [97]
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2. 26/20 = 130%
7 0
3 years ago
1. Use the points in the diagram to name the figure. <br> C<br> D<br> is is 8 IA<br> DC
IgorC [24]

Answer:

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5 0
3 years ago
Based on the z-statistic of –2.5 for the sample data, should the null hypothesis be rejected? Yes, because the z-statistic lies
ivann1987 [24]

Answer:

No, because z-statistics lies within the critical region.

Step-by-step explanation:

The z value is calculated to determine whether the null hypothesis should be accepted or rejected. The null hypothesis is rejected when the z-value lies outside the critical region. The critical region is determined by the sample data.

6 0
3 years ago
Read 2 more answers
A box holds 8 red and 4 blue beads. Three beads are taken from the box and not replaced. Determine: The probability that all thr
zubka84 [21]

Answer: \dfrac{14}{55}, \dfrac{13}{55}

Step-by-step explanation:

Given

There are 8 red and 4 blue beads

Three beads were taken from the box

(i)No of ways of choosing three red beads out of 12 beads is ^8C_3

Total no of ways ^{12}C_3

Probability is

P=\dfrac{^8C_3}{^{12}C_3}=\dfrac{8\times 7\times 6}{12\times 11\times 10}=\dfrac{2\times 7}{11\times 5}\\P=\dfrac{14}{55}

(ii)atleast two beads refers to minimum two beads

There can be two possibilities (2B,1 R), (3B, 0R)

\Rightarrow P=\dfrac{^4C_2\times ^8C_1}{^{12}C_3}+\dfrac{^4C_3}{^{12}C_3}\\\\\Rightarrow P=\dfrac{6\times 8}{220}+\dfrac{4}{220}=\dfrac{52}{220}\\\\\Rightarrow P=\dfrac{13}{55}

6 0
3 years ago
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