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Ede4ka [16]
3 years ago
11

I went to the grocery store the other day and there were no price tags on anything! I bought 3 loaves of bread and 1 milk jug an

d when the cashier scanned it, the total was $9. Then I saw the person behind me bought 1 loaf of bread and 2 milk jugs. Her total was $5.50. I walked out: how much does one loaf of bread cost?
Mathematics
1 answer:
Lostsunrise [7]3 years ago
8 0
To solve this problem, set up and solve a system of equations.  The variables b and m will represent a bread loaf and milk jug, respectively:

\left \{ {{3b + m = 9} \atop {b + 2m =5.5}} \right.

I would solve using substitution.  Take one of the equations and set it equal to one of the variables, for example:

3b + m =9 \\ m = 9 -3b

Now, plug this into the other equation for m and solve for b:

b + 2m =5.5 \\ b + 2(9-3b)= 5.5 \\ b + 18 - 6b = 5.5 \\ -5b + 18 = 5.5 \\ -5b = -12.5 \\ b = 2.5

We now know that a loaf of bread costs $2.50.  Plug this value in for b in the first equation and solve for m:

3b + m = 9 \\ 3(2.5) + m = 9 \\ 7.5+m = 9 \\ m = 1.5

One jug of milk costs $1.50 and one loaf of bread costs $2.50.
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Solve 4x-2y=6 for x in terms for y
Elina [12.6K]

Answer:

x = 3 + y

Step-by-step explanation:

4x - 2y = 6

~Add 2y to both sides

2x = 6 + 2y

~Divide 2 to everything

x = 3 + y

Best of Luck!

5 0
2 years ago
6. Write the explicit formula for the geometric sequence of the height of the ball on the 10th bounce.First bounce: 54 inchescom
Nat2105 [25]

The height of the ball initially is 54 inches. The common ratio has been provided which is 0.750.

The explicit formular for the geometric sequence is shown as;

\begin{gathered} ar^{n-1} \\ \text{Where a is the first term in the sequence,} \\ r\text{ is the common ratio betw}een\text{ every successive term} \\ \text{And n is the nth term} \\ \text{The explicit formular for the height of the ball on the 10th bounce is;} \\ ar^{n-1} \\ 54\times(0.750)^{10-1} \\ 54\times(0.750)^9 \\ 54\times0.07508468\ldots \\ 4.0545 \end{gathered}

The explicit formula is shown as;

54 x (0.750)^9

The rest is simply to solve for the height on the 10th bounce

5 0
1 year ago
Can anyone answer this problem in their head. No calculator or paper.
rodikova [14]
7,393,420,225 is the answer tk the problem without using calculator.
7 0
3 years ago
. (0.5 point) We simulate the operations of a call center that opens from 8am to 6pm for 20 days. The daily average call waiting
SashulF [63]

Answer:

The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

The change in the call waiting time is not statistically significant

Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27

From the data we have;

The mean waiting time before the downsize, \overline x_1 = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, \overline x_2 = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20  - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; t_{\alpha /2} = ±2.026192

The t-confidence interval is given as follows;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore;

\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

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Answer:

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