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Vaselesa [24]
2 years ago
5

PLS help will give brainliest!!! 50 Points!!

Mathematics
1 answer:
lbvjy [14]2 years ago
6 0

Answer:

Yes they are equal to each other.

Step-by-step explanation:

Sorry i can't use links

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Suppose a study estimated that 62% of the residents of a community (with an
kati45 [8]

Answer:

D. 55%

Step-by-step explanation:

The error range ± 9% means that the range is 9% below 62% or 9% above 62%.

The range is:

(62% - 9%) to (62% + 9%)

53% to 71%.

The only answer between 53% and 71% is 55%.

Answers A, B, C are all greater than 71%.

5 0
3 years ago
The graph of y= 3x-2 is translated up 5 units. What is the new equation of the graph.
algol13

Answer:

y = 3x + 3

Step-by-step explanation:

If the entire line, or rather, the graph, is translated upwards by 5 units; the slope stays the same, but the y-intercept rises by 5 as well. Thus, the equation of the new graph would be;

y = 3x (-2+5)

y = 3x + 3

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2 years ago
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Which term best describes the graph of yy = x - 1
dimaraw [331]

Answer:

C

Step-by-step explanation:

Its C because its x (a variable) and is costanly changing

4 0
3 years ago
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

8 0
3 years ago
FREE GIVE AWAY IN 5 MINUTES
Olegator [25]

Answer:

Thanks.......... follow me

8 0
2 years ago
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