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madreJ [45]
3 years ago
12

Zeke spent 12 minutes on the phone while routing 4 phone calls. In all, how many phone

Mathematics
1 answer:
Keith_Richards [23]3 years ago
3 0
5 because
12/4 = 15/x
X=5
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Need help it seems so simple but i dont know what it is
wel
3,5 ?
Hope this helps!
4 0
4 years ago
Read 2 more answers
Find dy/dx by implicit differentiation and evaluate the derivative at the given point.xy = 12, (-4, -3)
mojhsa [17]
Xy=12
xdy/dx + y = 12
xdy/dx = 12 - y
dy/dx= (12-y) /x
dy/dx | x=-4 ,y=-3 = (12-(-3))/(-4)
= (12+3)/-4 = -15/4
4 0
4 years ago
In details<br> Don't spam
Scrat [10]

Answer:

a = -4.

Step-by-step explanation:

25(√5)^a * (√5) ^3  = 5√5

25*5^a/2 * 5^3/2 = 5*5^1/2

25*5^a/2  = 5*5^1/2 / 5^3/2

25*5^a/2  = 5^1 *5^1/2 / 5^3/2

25*5^a/2 = 5^3/2 / 5^3/2 = 1

5^a/2 =  1/25 = 5^-2

The bases are equal, so

a/2 = -2

a = -4.

7 0
4 years ago
Read 2 more answers
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
How do i add 567,890 567,096
marissa [1.9K]
567,890
+567,096
________
1,134,986
3 0
3 years ago
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