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oksian1 [2.3K]
3 years ago
10

Thomas bought 4 shirts from a store. The price of each shirt was the same. After he bought the shirts, his account balance showe

d a change of −$89.08. What would have been the change to Thomas's account balance had he bought only 1 shirt from the store?
Mathematics
2 answers:
ludmilkaskok [199]3 years ago
8 0
$89.08 divided by 4 is $22.27, aka the price of one shirt
Helga [31]3 years ago
5 0

Answer:

Step-by-step explanation:

negitive $22.27

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A pole that is 3m tall casts a shadow that is 1.3m long. At the same time, a nearby building casts a shadow that is 41.75m long.
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Answer:We're looking for a,b such that

\dfrac{9x-20}{(x+6)^2}=\dfrac a{x+6}+\dfrac b{(x+6)^2}

Step-by-step explanation:

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7 times as much as the sum of 1/3 and 4/5
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7( \frac{1}{3} +  \frac{4}{5} )

=7( \frac{1(5)}{3(5)} +  \frac{4(3)}{5(3)} )

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4 0
3 years ago
Read 2 more answers
In a large population of college-educated adults, the mean IQ is 118 with a standard deviation of 20. Suppose 200 adults from th
cricket20 [7]

Answer: 0.0793

Step-by-step explanation:

Let the IQ of the educated adults be X then;

Assume X follows a normal distribution with mean 118 and standard deviation of 20.

This is a sampling question with sample size, n =200

To find the probability that the sample mean IQ is greater than 120:

P(X > 120) = 1 - P(X < 120)

Standardize the mean IQ using the sampling formula : Z = (X - μ) / σ/sqrt n

Where;  X = sample mean IQ; μ =population mean IQ; σ = population standard deviation and n = sample size

Therefore, P(X>120) = 1 - P(Z < (120 - 118)/20/sqrt 200)

                                 = 1 - P(Z< 1.41)

The P(Z<1.41) can then be obtained from the Z tables and the value is 0.9207

Thus; P(X< 120) = 1 - 0.9207

                          = 0.0793

3 0
3 years ago
Two linear equations are shown. A coordinate grid with 2 lines. The first line is labeled y equals StartFraction one-third EndFr
Lorico [155]

Answer:

(7,\frac{13}{3})

Step-by-step explanation:

we have

<em>The equation of the first line</em>

y=\frac{1}{3}x+2 ------> equation A

<em>The equation of the second line</em>

y=\frac{4}{3}x-5 ------> equation B

Solve the system of equations by elimination

Multiply equation A by -4 both sides

(-4)y=(-4)(\frac{1}{3}x+2)

-4y=-\frac{4}{3}x-8 --------> equation C

Adds equation B and equation C

y=\frac{4}{3}x-5\\-4y=-\frac{4}{3}x-8\\--------\\y-4y=-5-8\\-3y=-13\\y=\frac{13}{3}

<em>Find the value of x</em>

substitute the value of y

\frac{13}{3}=\frac{1}{3}x+2

\frac{1}{3}x=\frac{13}{3}-2

Multiply by 3 both sides

x=13-6

x=7

therefore

The solution to the system of equations is the point (7,\frac{13}{3})

5 0
3 years ago
Read 2 more answers
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