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2 years ago

How many solutions exist for the given equation? 12x + 1 = 3(4x + 1) – 2

Mathematics

1 answer:

lubasha [3.4K]2 years ago

6 0

Answer:

3=3 infinity many solutions

Step-by-step explanation:

12x+1=3(4x+1)-1

3x4x 3x1

12x+1=12x+3-2

+2 +2

12x+3=12x+3

-12x -12x

3=3

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marshall27 [118]

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2 years ago

Adam is going to make concrete. He is going to use 180 kg of cement. 375 kg of sand. 1090 kg of stone. Cement, sand and stone ar

hichkok12 [17]

Answer:

2 bags of stone

Step-by-step explanation:

Lets do one at a time. Cement first.

he needs 180 kg of cement.

1 bag is 25 kg, so in total he needs 8 because 7 would get him 175, so he would need just a little bit more. And I assume you can't but part of a bag he needs to but one full more bag. If he already has 8 then he doesn't need any more.

Needs 375 kg of sand

Each bag has 22.5 kg inside

he needs 17 bags total of sand because 16 would only give 360

Since he already has 19 bags he has 427.5 kg of sand so he doesn't need any more.

Needs 1090 kg of stone

Each bag has 50 kg of stone

He needs 22 bags since 21 would just give him 1050 kg

He has 20 so he needs two more bags

So in total he needs two bags of stone. Let me know if there was anything you didn't understand.

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3 years ago

Is what I did right?

yuradex [85]

Yes you got the problem right

3 0

2 years ago

Read 2 more answers

The expression d/t can be used to find the average speed of an object that travel a distance d in time t. What is sc cars averag

Firlakuza [10]

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3 years ago

A coin is thrown independently 10 times to test the hypothesis that the probability of heads is 0.5 versus the alternative that

mafiozo [28]

Answer:

(a) The significance level of the test is 0.002.

(b) The power of the test is 0.3487.

Step-by-step explanation:

We are given that a coin is thrown independently 10 times to test the hypothesis that the probability of heads is 0.5 versus the alternative that the probability is not 0.5.

The test rejects the null hypothesis if either 0 or 10 heads are observed.

Let p = <u><em>probability of obtaining head.</em></u>

So, Null Hypothesis, $H_0$ : p = 0.5

Alternate Hypothesis, $H_A$ : p $\neq$ 0.5

(a) The significance level of the test which is represented by $\alpha$ is the probability of Type I error.

Type I error states the probability of rejecting the null hypothesis given the fact that the null hypothesis is true.

Here, the probability of rejecting the null hypothesis means we obtain the probability of observing either 0 or 10 heads, that is;

P(Type I error) = $\alpha$

P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true) = $\alpha$

Also, the event of obtaining heads when a coin is thrown 10 times can be considered as a binomial experiment.

So, X ~ Binom(n = 10, p = 0.5)

P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true) = $\alpha$

$\binom{10}{0}\times 0.5^{0} \times (1-0.5)^{10-0} +\binom{10}{10}\times 0.5^{10} \times (1-0.5)^{10-10}$ = $\alpha$

$(1\times 1\times 0.5^{10}) +(1 \times 0.5^{10} \times 0.5^{0})$ = $\alpha$

$\alpha$ = 0.0019

So, the significance level of the test is 0.002.

(b) It is stated that the probability of heads is 0.1, and we have to find the power of the test.

Here the Type II error is used which states the probability of accepting the null hypothesis given the fact that the null hypothesis is false.

Also, the power of the test is represented by (1 - $\beta$ ).

So, here, X ~ Binom(n = 10, p = 0.1)

$1-\beta$ = P(X = 0/ $H_0$ is true) + P(X = 10/ $H_0$ is true)

$1-\beta$ = $\binom{10}{0}\times 0.1^{0} \times (1-0.1)^{10-0} +\binom{10}{10}\times 0.1^{10} \times (1-0.1)^{10-10}$

$1-\beta$ = $(1\times 1\times 0.9^{10}) +(1 \times 0.1^{10} \times 0.9^{0})$

$1-\beta$ = 0.3487

Hence, the power of the test is 0.3487.

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2 years ago