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gregori [183]
3 years ago
12

Clair went shopping for school clothes.If the items cost $150 and he receives a 10% discount,what is the final price of this pur

chase?
Mathematics
1 answer:
Illusion [34]3 years ago
6 0
The final price is $135 , because
.10 × 150 = 15
150 - 15 = 135
You might be interested in
Pls, look at picture for the question.
Lady bird [3.3K]

point D,C,A...... ......

8 0
3 years ago
Read 2 more answers
Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time  t > 0 is \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t

where, \text{R}_i_n = concentration of salt in the inflow \times input rate of brine solution

and \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

So, \text{R}_i_n = 4 lb/gal \times 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

                                                = 3 gal/min - 2 gal/min

                                                = 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

             = \frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min} = \frac{2A(t)}{500+t} \text{ lb/min }

Now, the differential equation for the amount of salt A(t) in the tank at a time  t > 0 is given by;

= \frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }

or \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

4 0
3 years ago
The mayor is interested in finding a 98% confidence interval for the mean number of pounds of trash per person per week that is
saul85 [17]

Solution :

Given :

Sample mean, $\overline X = 34.2$

Sample size, n = 129

Sample standard deviation, s = 8.2

a. Since the population standard deviation is unknown, therefore, we use the t-distribution.

b. Now for 95% confidence level,

   α = 0.05, α/2 = 0.025

  From the t tables, T.INV.2T(α, degree of freedom), we find the t value as

  t =T.INV.2T(0.05, 128) = 2.34

  Taking the positive value of t, we get  

  Confidence interval is ,

  $\overline X \pm t \times \frac{s}{\sqrt n}$

 $34.2 \pm 2.34 \times \frac{8.2}{\sqrt {129}}$

 (32.52, 35.8)

 95% confidence interval is  (32.52, 35.8)

So with $95 \%$ confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.

c. About $95 \%$ of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about $5 \%$ that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.  

4 0
3 years ago
Help me asap! I will mark you brainly and a heart tysm!<br><br><br> :D
RideAnS [48]
The answer is A) -4/25 because when you multiply 8/15 and -3/10, you can simplify the expression to 4/5 times -1/5, which equals -4/25.  
7 0
3 years ago
Read 2 more answers
WILL MARK BRAINLIEST! PLEASE HELP!<br> Write 206 in base 7
Anna71 [15]

Answer:

413_7

Step-by-step explanation:

Convert the following to base 7:

206_10

Hint: | Starting with zero, raise 7 to increasingly larger integer powers until the result exceeds 206.

Determine the powers of 7 that will be used as the places of the digits in the base-7 representation of 206:

Power | \!\(\*SuperscriptBox[\(Base\), \(Power\)]\) | Place value

3 | 7^3 | 343

2 | 7^2 | 49

1 | 7^1 | 7

0 | 7^0 | 1

Hint: | The powers of 7 (in ascending order) are associated with the places from right to left.

Label each place of the base-7 representation of 206 with the appropriate power of 7:

Place | | | 7^2 | 7^1 | 7^0 |  

| | | ↓ | ↓ | ↓ |  

206_10 | = | ( | __ | __ | __ | )_(_7)

Hint: | Divide 206 by 7 and find the remainder. The remainder is the first digit from the right.

Determine the value of the first digit from the right of 206 in base 7:

206/7=29 with remainder 3

Place | | | 7^2 | 7^1 | 7^0 |  

| | | ↓ | ↓ | ↓ |  

206_10 | = | ( | __ | __ | 3 | )_(_7)

Hint: | Divide the whole number part of the previous quotient, 29, by 7 and find the remainder. The remainder is the next digit.

Determine the value of the next digit from the right of 206 in base 7:

29/7=4 with remainder 1

Place | | | 7^2 | 7^1 | 7^0 |  

| | | ↓ | ↓ | ↓ |  

206_10 | = | ( | __ | 1 | 3 | )_(_7)

Hint: | Divide the whole number part of the previous quotient, 4, by 7 and find the remainder. The remainder is the last digit.

Determine the value of the last remaining digit of 206 in base 7:

4/7=0 with remainder 4

Place | | | 7^2 | 7^1 | 7^0 |  

| | | ↓ | ↓ | ↓ |  

206_10 | = | ( | 4 | 1 | 3 | )_(_7)

Hint: | Express 206_10 in base 7.

The number 206_10 is equivalent to 413_7 in base 7.

Answer: 206_10 =413_7

5 0
2 years ago
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