Help please..............

Please help me pleasse I really really need help

snow_lady [41]

Answer:

$8.88

Step-by-step explanation:

24yds needed = 864 inches

864/150 = 5.76 so he'll need 6 spools

6x 1.48 = $8.88

5 0

3 years ago

Read 2 more answers

The sale price of the sword was $20. It had been marked down 23% for the sale. What was the original

Viefleur [7K]

Answer:

$24.60

Step-by-step explanation:

8 0

3 years ago

Find the vertex of this parabola: y=-x^2+2x-7

Luden [163]

y=−x2+2x−7y=-x2+2x-7Complete the square on the right side of the equation.Tap for more steps...−(x−1)2−6-(x-1)2-6Reorder the right side of the equation to match the vertex form of a parabola.y=−(x−1)2−6y=-(x-1)2-6Use the vertex form, y=a(x−h)2+ky=a(x-h)2+k, to determine the values of aa, hh, and kk.a=−1a=-1h=1h=1k=−6k=-6Find the vertex (h,k)(h,k).(1,−6)(1,-6)

6 0

4 years ago

The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

(c) The probability that at least one contains high levels of contamination is

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

6 0

3 years ago

Can someone do this for me I was busy helping around for my mom and myself please it's due today!!!!!!

Fed [463]

Answer:

first discount for Q7 is $54.3915

the next discount is $48.95235

for Q8 is

Step-by-step explanation:

you are a very nice person for helping your mom. good luck for this!

ok let's work on question 7 first.

$63.99 is the cost of a skateboard.

15% of 63.99 = 10% + 5%

10% of 63.99 = 6.399 (move the decimal one place to the left)

5% of 63.99 = 1/2 of 6.399

1/2 x 6.399 = 3.1995

6.399 + 3.1995 = 9.5985

ok, so that's 15% of 63.99

now we need to take away 9.5985 from 63.99

63.99 - 9.5985 = 54.3915

the discount price of the skateboard is $54.3915

a further 10% is $5.43915

so take off that from the discount,

54.3915 - 5.43915 = 48.95235

the final price is $48.95235

QUESTION 8

shop A = 6.95 x 40

6.95 x 40 = $278

shop B = 20% of 325

20/100 * 325 = 65

325 - 65 = 260

therefore, shop B is cheaper.

hope this helped, good luck!

4 0

3 years ago