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1 year ago

What is the solution to –2(5y – 5) – 3y ≤ –7y + 22?

2 answers:

6 0

-2(5y - 5) - 3y < = -7y + 22
-10y + 10 - 3y < = -7y + 22
-13y + 10 < = -7y + 22
-13y + 7y < = 22 - 10
-6y < = 12
y > = -12/6
y > = -2....answer B

slavikrds [6]1 year ago

4 0

Answer:

Option: B is the correct answer.

y≥ -2

Step-by-step explanation:

We are asked to find the solution of the inequality:

–2(5y – 5) – 3y ≤ –7y + 22.

We will solve for the inequality as:

-2×5y-2×(-5)-3y ≤ -7y+22 (on solving for the parantheses in the left hand side of the inequality)

-10y+10-3y ≤ -7y+22

⇒ -10y+7y-3y ≤ 22-10 (on combining the like terms)

⇒ -6y ≤ 12

⇒ y≥ -2 (on dividing both side by -6 also the sign of the inequality changes as the negative sign is treated on both the side)

Hence option: B is the correct answer.

y ≥ -2.

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A bank account earned 3.5% continuously compounded annual interest. After the initial deposit, no deposits or withdrawals were m

aalyn [17]

This question is incomplete. Below is the complete question:

A bank account earned 3.5% continuously compounded annual interest. After the initial deposit, no deposits or withdrawals were made. At the end of an 8 year period, the balance in the account was $13231.30. what is the amount of the initial deposit?

Answer: The initial deposit is $10,001

Step-by-step explanation:

To solve this we need to utilize the continuous compounding interest formula:-

Fv = Pv × e^(i × t)

Where, Fv = the future value

Pv = present value

i = the interest rate

t = the time in years

e = a mathematical constant that is usually approximated by 2.7183

In this case, the Fv = 13,231.30

i = 3.5% or 0.035

t = 8 years

e = 2.7183

Pv = ? (The unknown variable).

13,231.30 = Pv × [2.7183^(0.035 × 8)]

13,231.30 = Pv × [2.7183^(0.28)]

13,231.30 = Pv × 1.323

Pv = 13,231.30/1.323

Pv = $10000.98

= $10,001

Therefore the initial deposit is $10,001

8 0

1 year ago

The solution to the given system of linear equations lies in which quadrant?

Rina8888 [55]

X-3y=6
x+y=2
this is an substitution problem
so first you can do is rewrite the problem by subjection one variable
x=3y+6
then substitute this in the other proble
x+y=2
(3y+6)+y=2
4y+6=2
4y=2-6
4y=-4
y=-1
then substitute the no. in the original equation.
x=3y+6
x=3(-1)+6
x=-3+6
x=3
now you got the intercepts and you draw the line and check.
it's in the IV quadrant

5 0

1 year ago

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The manager of a symphony in a large city wants to investigate music preferences for adults and students in the city. Let pA rep

monitta

Answer:

Answer E

Step-by-step explanation

The statement gives a probability of approximately 0.022 for the difference in sample proportions, pˆA−pˆS, being greater than 0.

6 0

1 year ago

Let L be the line with parametric equations x=2+t, y=1-t, z=1+3t.Let v=(1,2,0).Find vectors w1 and w2 such that v= w1 + w2, and

charle [14.2K]

Answer:

w1 = (-1/11, 1/11, -3/11)

w2 = (12/11, 21/11, 3/11)

Step-by-step explanation:

Direction ratio of w1 = Direction ratio of L (because parallel) = K+(1, -1, 3)

Let <a,b,c> be direction ratio of w2.

Then, <a,b,c>. <1,-1,3> = 0

a-b+3c = 0

v = w1 + w2

(1, 2, 0) = k(1, -1, 3) + (a, b, c)

a + k = 1

b - K = 2

c - 3k = 0

Solving 4 equations, a = 12/11, b= 21/11, c = 3/11, k=-1/11

So, w1 = -1/11(1, -1 ,3) = (-1/11, 1/11, -3/11)

w2 = (12/11, 21/11, 3/11)

6 0

1 year ago

Two different samples will be taken from the same population of test scores where the population mean and standard deviation are

Alenkinab [10]

Answer:

The sample consisting of 64 data values would give a greater precision.

Step-by-step explanation:

The width of a (1 - α)% confidence interval for population mean μ is:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{n}}$

So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (n).

That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.

Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.

The two sample sizes are:

n₁ = 25

n₂ = 64

The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.

Width for n = 25:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{25}}=\frac{1}{5}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]$

Width for n = 64:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]$

Thus, the sample consisting of 64 data values would give a greater precision

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1 year ago

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