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3 years ago

What is the answer to this equation 2/3x +7= 5

Mathematics

2 answers:

nadezda [96]3 years ago

5 0

Answer:

x = -3

Step-by-step explanation:

$\frac{2}{3} x + 7 = 5\\\frac{2}{3} x = 5 - 7\\\frac{2}{3} x = -2\\x = -2 / \frac{2}{3}\\x = -2 * \frac{3}{2}\\x = -6/2\\x = -3$

kow [346]3 years ago

5 0

Answer:

x= -3

Step-by-step explanation:

2/3x+7=5

subtract 7 to both side (2/3x +7 -7) = 5-7

(2/3x +7 -7 = 5-7)= 2/3x= -2

then multiple by 3/2 on both sides 2/3x times 3/2 = -2 times 3/2

(2/3x times 3/2 = -2 times 3/2)= (x= -3)

3/2 times -2= -6/2 or -3

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How to do question five

-BARSIC- [3]

I believe the answer is 46.
Remember "PEMDAS" when doing equations like these :)

8 0

3 years ago

Ann made a scale drawing of a theater. The scale she used was 1 inch : 7 feet. The stage is

hjlf

Answer:

4 in

Step-by-step explanation:

<u>Use ratios and solve:</u>

1 in ÷ 7 ft = x in ÷ 28 ft
x = 28/7 in
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4 0

3 years ago

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q = -t/100 + c

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

4 years ago

Assume the general population gets an average of 7 hours of sleep per night. You randomly select 45 college students and survey

tekilochka [14]

Answer:

a) ii. This is a left-tailed test.

b) -1.59

c) -1.301

d) i. reject null hypothesis

e) Option i) The data supports the claim that college students get less sleep than the general population.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 7 hours

Sample mean, $\bar{x}$ = 6.87 hours

Sample size, n = 45

Alpha, α = 0.10

Sample standard deviation, s = 0.55 hours

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 7\text{ hours}\\H_A: \mu < 7\text{ hours}$

a) We use one-tailed(left) t test to perform this hypothesis.

b) Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{6.87 - 7}{\frac{0.55}{\sqrt{45}} } =-1.59$

c) Now,

$t_{critical} \text{ at 0.10 level of significance, 44 degree of freedom } = -1.301$

Since,

$t_{stat} < t_{critical}$

d) We fail to accept the null hypothesis and reject it.

We accept the alternate hypothesis and conclude that mean number of hours of sleep for all college students is less than 7 hours.

e) Option i) The data supports the claim that college students get less sleep than the general population.

8 0

3 years ago

Help please very much

Norma-Jean [14]

Answer:

Step-by-step explanation:

3 0

3 years ago