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Tanzania [10]
3 years ago
6

When we take a census . We attempt to collect data from a. A stratified random sample b. Every individual chosen in a simple ran

dom sample . c. Every individual in the population D. A voluntary response sample E. A convenience sample
Mathematics
1 answer:
In-s [12.5K]3 years ago
4 0

Answer:c

Step-by-step explanation:

Be

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How can these models be used to find the sum 1.8 + 1.56?
enyata [817]

Answer:

The sum of  1.8 and 1.56 is 3.36 according to the model

Sum of decimals

Given the expression  1.8 + 1.56

Using the partial sum method

1.8 + 1.56 = (1.0 + 0.8) + (1.0+0.56)

Regroup as whole and decimal number

1.8 + 1.56 = (1.0 + 1.0) + (0.8 +0.56)

1.8 + 1.56 = 2.0 + 1.36

1.8 + 1.56 = 3.36

Hence the sum of  1.8 and 1.56 is 3.36 according to the model

3 0
2 years ago
Identify steps. to solve -3 + x =11 <br>​
podryga [215]
-3 + x = 11
+3 +3
x =14
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3 years ago
WILL MARK BRANLIEST
Naily [24]

Answer:

Brand b has a better unit price.

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3 years ago
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Solve for the given variable 7x+5y=6 for y
olga_2 [115]

Answer:

y = (6/5) - (7/5) x

Step-by-step explanation:

goal: to rearrange the equation such that it takes the form:

y = f(x) ; where f(x) is an expression in terms of x

Given:

7x+5y=6   subtract 7x from both sides

5y = 6 - 7x   (divide both sides by 5)

y = (6-7x) / 5

y = (6/5) - (7/5) x

4 0
2 years ago
A game of chance involves rolling an unevenly balanced 4-sided die. The probability that a roll comes up 1 is 0.21, the probabil
WITCHER [35]

Answer:

2.65 dollars

Step-by-step explanation:

The expected value for a discrete variable is calculated as:

E(x)=x1P(x1) + x2P(x2) + ... + xnP(xn)

Where x1, x2, ... , xn are the possibles values of the variable and P(x1), P(x2), ... , P(xn) are the probabilities of x1, x2, ... , xn respectively.

In this case, the roll can comes up 1, 2, 3 or 4 and you can win 1, 2, 3 or 4 dollars respectively. So, taking into account that they are mutually exclusive events, the probability that the player win 1, 2, 3 or 4 dollars is:

P(1) = 0.21

P(2) = P(1∪2) - P(1) = 0.42 - 0.21 = 0.21

P(3) = P(2∪3) - P(2) = 0.51 - 0.21 = 0.3

P(4) = 1 - P(1) - P(2) - P(3) = 1 - 0.21 - 0.21 - 0.3 = 0.28

Therefore, If you win the amount that appears on the die the expected winning are:

E(x) = $1P(1) + $2P(2) + $3P(3) + $4P(4)

E(x) = $1(0.21) + $2(0.21) + $3(0.3) + $4(0.28)

E(x) = $2.65

4 0
3 years ago
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