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SCORPION-xisa [38]
2 years ago
6

Jesse needs to wrap the present shown below. How much wrapping paper would he need?

Mathematics
1 answer:
maxonik [38]2 years ago
4 0
6in would be the answer
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larisa86 [58]

Answer:

14, 16, 18, 20

Step-by-step explanation:

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3 years ago
-. For each equation below, determine whether
Andre45 [30]

Answer:

Step-by-step explanation:

The first answer a, is not a linear function because of the first term, 2√(x), this is basically 2(x^(1/2)) which means that it is not linear

B is linear because there are no exponents in any of the variables.

C is not linear because it forms a circle when graphed

D is linear because there are no exponents in any of the variables.

E is linear because there are no exponents in angy of the variables.

F is linear because there are no exponents in any of the variables.

G is not linear because it is a quadratic equation.

7 0
3 years ago
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Find the sales tax on each item<br>19) tennis racket<br>cost: $59.98;<br>sales tax: 6%<br>​
valentinak56 [21]

Before Tax Price: $59.98

Sale Tax: 6.00% or $3.60

After Tax Price: $63.58

6 0
3 years ago
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What is the scale factor for pre-image IJKH to image MNOL?
solniwko [45]

Answer:

Step-by-step explanation:

Given are two rectangles MNOL and IJKH

MNOL is the image and IJKH is the preimage.

OBviously similarity is preserved between these.

Hence sides will be proportional

i.e. l1/l2 = w1/w2 = Scale factor k

This can also be written as

Scale factor square = k^2=\sqrt{ \frac{areaI}{area2} }

Scale factor = square root of ratio of rectangle areas

Find ratio as

\frac{375}{135} =\frac{25}{9}

k = Scale factor = square root of the above fraction

=\frac{5}{3}

5 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
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