Subtract the linear expressions.

Is (2, 5) a solution to this system of equations? 6x + y = 17 3x + 14y = 16 yes or no

Anarel [89]

Answer:

Putting the value of x= 2 in the first equation,

6(2)+y= 17

y= 17-12 = 5

This value is same as the value for y given in the question.

Therefore, it satisfies the equation.

Again, putting the value of x= 2 in the second equation,

3(2)+14y= 16

14y= 16-6 = 10

y= 10/14 = 5/7

It doesn't satisfies the 2nd equation

Hence, (2,5) is not the solution to this system.

5 0

3 years ago

Speed test cost $5 each and large one-topping pizzas cost $6 each write an equation that represents the total cost t of c large

yKpoI14uk [10]

Answer:

Equation t = 5c + 6d represents total cost t of c large pizzas and d large one-topping pizzas

Solution:

Given that

Cost of large pizza = $5 each

Cost of large one-topping pizza = $6 each

Need to write the equation that represents total cost t of c large pizzas and d large one topping pizzas.

From given information

Cost of 1 large pizza = 5

So cost of c large pizzas = c x 5 = 5c

Cost of 1 large one-topping pizza = 6

So cost of d large one-topping pizzas =d x 6 = 6d

Total cost t = cost of c large pizzas + cost of d large one-topping pizzas

=> t = 5c + 6d

Hence equation t = 5c + 6d represents total cost t of c large pizzas and d large one-topping pizzas.

3 0

3 years ago

In the electric currect flow, it is found that the resistance (measured in the units called ohms) offered by a fixed length o

Morgarella [4.7K]

Let x be the resistance of the wire with the larger diameter.
x = (0.456 x (0.01)^2)/(0.0401)^2 = 0.0284 ohms
The answer is 0.0284 ohms.

7 0

2 years ago

Find the diameter of each circle.

Vedmedyk [2.9K]

Answer:

Diameter Length: ( About ) 5.4 km; Option B

Step-by-step explanation:

~ Let us apply the Area of the Circle formula πr^2, where r ⇒ radius of the circle ~

1. We are given that the area of the circle is 22.9 km^2, so let us substitute that value into the area of the circle formula, solving for r ( radius ) ⇒ 22.9 = π * r^2 ⇒ r^2 = 22.9/π ⇒ r^2 = 7.28929639361.... ⇒

radius = ( About ) 2.7

2. The diameter would thus be 2 times that of the radius by definition, and thus is: 2.7 * 2 ⇒ ( About ) 5.4 km

Diameter Length: ( About ) 5.4 km

5 0

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 = probability of obtaining head.

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.

3 0

3 years ago