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adelina 88 [10]
4 years ago
7

A blender that normally costs $130 is on sale for 12% off. What is the final cost of the blender if there is also an 8% sales ta

x? Round your answer to the nearest cent.
Mathematics
1 answer:
polet [3.4K]4 years ago
8 0

Answer:

$105.25

Step-by-step explanation:

First we need to know to price of the blender after the sale, so we are going to pay the 88% of the blender's price so the final price without the sale tax is going to be

\$130\times .88=\$ 114.4      (1)

Knowing this price we need to calculate the 8% of the total price

\$ 114.4 \times .08=\$ 9.152    (2)

We substract the result (2) from (1)

\$ 114.4 - \$ 9.152 = 105.248

Rounding this number to nearest cent the final result is:

$105.25

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Estimate the difference. Round each number to the nearest whole number, then subtract.
Y_Kistochka [10]

Answer:

nothing is the difference

3 0
3 years ago
The grade appeal process at a university requires that a jury be structured by selecting eight individuals randomly from a pool
anygoal [31]

Answer: The probability of selecting a jury of all​ faculty=0.000071

The probability of selecting a jury of six students and two two ​faculty=0.3667


Step-by-step explanation:

Given: The number of students = 9

The number of faculty members=11

The total number of ways of selecting jury of eight individuals=^{20}C_8=\frac{20!}{(20-8)!\times8!}=125970

The number of ways of selecting jury of all faculty=^9C_8=\frac{9!}{8!(9-8)!}=9

The probability of selecting a jury of all​ faculty=\frac{9}{125970}=0.000071

The number of ways of selecting jury of six students and two two ​faculty

=^9C_6\times ^{11}C_2=\frac{9!}{6!(9-6)!}\times\frac{11!}{2!\times(11-2)!}\\=84\times55=4620

Now, the probability of selecting a jury of six students and two two ​faculty

=\frac{4620}{125970}=0.03667


7 0
3 years ago
What is the value of x ?
Setler79 [48]

Answer:

19

Step-by-step explanation:

When you have a straight line with a transversal, the two angles are equal to 180 degrees

180=123+3x ---> subtract 123 to the other side

57=3x ---> divide by 3

x=57/3

x=19

3 0
3 years ago
6/7+3/5 simplest form
Sloan [31]

Answer:

\huge\boxed{\dfrac{6}{7}+\dfrac{3}{5}=\dfrac{51}{35}=1\dfrac{16}{35}}

Step-by-step explanation:

\dfrac{6}{7}+\dfrac{3}{5}\\\\\text{Find the}\ LCD-\text{Least Common Denominator}\\\\\text{List multiples of 7}:\ 0,\ 7,\ 14,\ 21,\ 28,\ \boxed{35},\ 42,\ ...\\\text{List multiples of 5:}\ 0,\ 5,\ 10,\ 15,\ 20,\ 25,\ 30,\ \boxed{35},\ 40,\ ...\\\\LCD(7,\ 5)=\boxed{35}\\\\35=7\cdot5

\dfrac{6}{7}+\dfrac{3}{5}=\dfrac{(6)(5)}{(7)(5)}+\dfrac{(7)(3)}{(7)(5)}=\dfrac{30}{35}+\dfrac{21}{35}=\dfrac{30+21}{35}=\dfrac{51}{35}=\dfrac{35+16}{35}=1\dfrac{16}{35}

5 0
3 years ago
Read 2 more answers
How many trucks carried only late variety?
murzikaleks [220]

9514 1404 393

Answer:

  • late only: 15
  • extra-late only: 24
  • one type: 43
  • total trucks: 105

Step-by-step explanation:

It works well when making a Venn diagram to start in the middle (6 carried all three), then work out.

For example, if 10 carried early and extra-late, then only 10-6 = 4 of those trucks carried just early and extra-late.

Similarly, if 30 carried early and late, and 4 more carried only early and extra-late, then 38-30-4 = 4 carried only early. In the attached, the "only" numbers for a single type are circled, to differentiate them from the "total" numbers for that type.

__

a) 15 trucks carried only late

b) 24 trucks carried only extra late

c) 4+15+24 = 43 trucks carried only one type

d) 38+67+56 -30-28-10 +6 +6 = 105 trucks in all went out

8 0
3 years ago
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