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3 years ago

How many ounces of a 40% alcohol solution must be mixed with 16 ounces of a 45% alcohol solution to make a 44% alcohol solution

Mathematics

1 answer:

goblinko [34]3 years ago

3 0

I'm pretty sure it's 4 ounces so it makes up 1/5 of 40% alcohol and 4/5 of 45% alcohol

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Solve 6 sec^2x-1= 7.

Dominik [7]

X=4

2x-1=7
Move the constant to the right - 2x=7+1
Calculate - 2x=8
Divide both sides - x=4

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3 years ago

Part 1 out of 2

Ainat [17]

Answer:

In June, the bakery gained $313.85

⇒ 2,205 – 1891.15 = 313.85

In July, the bakery gained $-77.33

⇒ 2,199 – 1807.82 = 391.18 ⇒ 313.85 (from June) – 391.18 = -77.33 gained OR 77.33 lost

In August, the bakery gained $55.65

⇒ 2209 – 1873.47 = 335.53 ⇒ 391.18 (from July) – 335.53 = 55.65 gained

Hope this helps! :)

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3 years ago

(6x^2 -x + 8) + (3x-4) Helppppp meeee

Alexus [3.1K]

Answer:

6x2+2x+4

Step-by-step explanation:

6x^2 has no like terms

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6x^2 + 2x + 4

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3 years ago

10+100000×10000×2000000

Mariulka [41]

Answer:

2e+15

Step-by-step explanation:

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3 years ago

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A market researcher for a provider of music player accessories wants to know the proportion of customers who own cars to assess

Luba_88 [7]

Answer:

Step-by-step explanation:

Hello!

To evaluate the market for a new car charger, a market researcher surveyed 600 customers and found out that 73% of them own a car.

The parameter of interest is the population proportion and the study variable is "the number of customers that own a car out of 600".

Using the Central Limit Theorem you can approximate the distribution of the sampling proportion to normal:

p'≈N(p; $\frac{p(1-p)}{n}$ )

Where p' is the sample proportion

p is the population proportion and the mean of the distribution

$\frac{p(1-p)}{n}$ is the variance of the distribution

Then de standard deviation of the samplig proportion distribution is $\sqrt{\frac{p(1-p)}{n}}$

You can estimate it using the sample proportion:

$\sqrt{\frac{p'(1-p')}{n}} = \sqrt{\frac{0.73*0.27}{600} }=0.0181$

Using the same sample proportion but a sample size of n=150, the estimated standard deviation is:

$\sqrt{\frac{p'(1-p')}{n}} = \sqrt{\frac{0.73*0.27}{150} }=0.0362$

I hope it helps!

6 0

3 years ago