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Rudik [331]
3 years ago
8

A furniture store is having a sale on couches. If a couch that originally costs $3400 is on sale for 15% off, what is the actual

cost of the couch?
Mathematics
1 answer:
vekshin13 years ago
8 0
To solve this one we need to find 15% of 3400 and then subtract it from 3400

to find percent we multiply so we'll multiply by 0.15

3400 times 0.15=510

3400-510=2890

then actual cost of the couch is $2890

I hope I've helped!
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IrinaVladis [17]

The perpendicular line will have an equation of y=3/4x-3/4

To find this, we first have to solve our equation for slope intercept form.

8x + 6y = -5 ----> subtract 8x

6y = -8x + -5 ----> divide by 6

y = -4/3x - 5/6.

So we know the slope of this equation to be -4/3. Since perpendicular lines have opposite and reciprocal slopes, we know we can simply flip the fraction and make it a negative to get the new slope of 3/4. Since B is the only option with that slope, we know it to be the correct answer.

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Domain: all real numbers
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Find the measures of the interior angles.
3241004551 [841]

Use angle sum property

  • x+68+x-24=180
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Please help with this question ASAP!
Naddik [55]

Answer:

8w : 3.8974342 ≈ 3.9 or 4 (hope it help)

Step-by-step explanation:

1w : 2

2w : 2 + 10% = 2.2

3w : 2.2 + 10% = 2.42

4w : 2.42 + 10% = 2.662

5w : 2.662 + 10% = 2.9282

6w : 2.9282 + 10% = 3.22102

7w : 3.22102 + 10% = 3.543122

8w : 3.543122 + 10% = 3.8974342

3.8974342 ≈ 3.9 or 4

6 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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