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adoni [48]
3 years ago
7

38+39+40+41+...+130+131

Mathematics
1 answer:
aliya0001 [1]3 years ago
6 0

Answer:

419

Step-by-step explanation:

the answer is 419 because 38+39=77+40=117+41=158+130=288+131=419

hope my math is correct =)

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Find the volume<br><br> 14.2 mm<br> 12 mm<br> 8.5 mm
irga5000 [103]

Answer:

1,448.4 mm.

Step-by-step explanation:

Volume is the length, width, and height multiplied together.

14.2 mm x 12 mm x 8.5 mm = 1,448.4 mm^2.

8 0
3 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
2 years ago
You bought a house in 1963 for $56,000. Since then, you have increased the value 6% per year. What is the value of your house in
7nadin3 [17]

Answer:

244,160

Step-by-step explanation:

8 0
3 years ago
Juanita is 10 years old her father is 4 times older than her when Juanita is 30 how old will her father be explain how you got y
almond37 [142]

answer: 60

age difference is 30 years

10x4=40

40-10=30

when she is 30 he will be 60 because 30(her age) + 30 (age difference) = 60

8 0
3 years ago
The recipe for Perfect Purple Water says, "Mix 8 ml of blue water with 3 ml of red water." Jada mixes 24 ml of blue water with 9
cluponka [151]

Answer:Jada will get same shade as Perfect Purple Water

Step-by-step explanation:

STEP 1

To get the right  recipe for Perfect purple water, every new mixture must have an equivalent ratio as the ideal mixture

The ideal mixture is given as

8ml  blue water and 3 ml red water.

Step 2

Andre mixes 16ml blue water with 9ml red water

\frac{8ml}{3ml} is not equivalent to \frac{16ml}{9ml}

Because dividing Andre's mixture  by a common value  will  not give  the equivalent ratio of the ideal recipe

\frac{16ml}{9ml}  \frac{/}{/}  \frac{3}{3} =\frac{5.3ml}{3ml}  

Jada mixes 24ml blue water with 9ml red water

\frac{8ml}{3ml} is equivalent to \frac{24ml}{9ml}

Because dividing Jada's mixture  by a common value  gives the equivalent ratio of the ideal recipe

\frac{24ml}{9ml}  \frac{/}{/}  \frac{3}{3} =\frac{8ml}{3ml}  

This shows that Jada got the right recipe and will get  same shade as Perfect Purple Water but in a higher quantity, 3 times the right recipe for  Perfect Purple Water

7 0
2 years ago
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