All categories

3 years ago

If x=7and =-2, determine the value of 6x-4y

Mathematics

2 answers:

SVEN [57.7K]3 years ago

8 0

Answer:

the answer is 50

Step-by-step explanation:

first you plug in the x and y in the equation, making it 6(7)=-4(-2) then, you multiply the numbers together, making the equation 42+8. finally, you add those numbers together, giving you the answer 50

slega [8]3 years ago

8 0

It's 50. Just multiply 6 times 7 and you get 42. Then you multiply 4 times -2 which is -8 . Then you subtract the products and you get 50. Your welcome.

You might be interested in

Find and simplify the difference quotient of the form<img src="https://tex.z-dn.net/?f=%5Cfra%5Cfrac%7Bf%28x%29-f%28a%29%7D%7Bx-

Ilia_Sergeevich [38]

Answer: 3x + 13

Step-by-step explanation:

$f(x)=3x^2+x\\f(a)=3a^2+a\\f(4)=3*4^2+4=52\\\\f(x)-f(4)=3x^2+x-52=(x-4)(3x+13)\\\\\dfrac{f(x)-f(4)}{x-4} =\dfrac{3x^2+x-52}{x-4} \\\\if\ x \neq 4\ then \\\\\dfrac{f(x)-f(4)}{x-4} =\dfrac{(x-4)(3x+13)}{x-4} \\\\=3x+13\\$

6 0

2 years ago

Read 2 more answers

A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe

d1i1m1o1n [39]

Answer:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

$df=n-1=15-1=14$

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

If we use a significance level of 0.05 we see that $p_v > \alpha$ and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation

$\bar X=5.63$ represent the sample mean

$s=1.61$ represent the standard deviation for the sample

$n=15$ sample size

$\mu_o =15/tex] represent the value that we want to test 
[tex]\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:

Null hypothesis: $\mu \leq 5.63$

Alternative hypothesis: $\mu > 5.63$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516$

Now we need to find the degrees of freedom for the t distribution given by:

$df=n-1=15-1=14$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{14}>1.516)=0.076$

3 0

3 years ago

Which of the following is equivalent to −2(4x + 8y) − 5(3q + 6p)?

Nadusha1986 [10]

Answer:

-8x + 16y -15q - 30p

Step-by-step explanation:

your just expanding brackets here

6 0

1 year ago

Read 2 more answers

Given the parent function f(x) = x², define the function g(x) that

Bas_tet [7]

Answer:

Part 1) $g(x)=-x^{2}$

Part 2) $g(x)=(x+3)^{2}$

Part 3) $g(x)=x^{2}+7$

Step-by-step explanation:

we have

The parent function is $f(x)=x^{2}$

This is a vertical parabola open upward with vertex at origin (0,0)

Part 1) Reflection over the x-axis

The rule of the reflection across the x=axis is equal to

(x,y) ------> (x,-y)

f(x) -----> g(x)

The function g(x) will be

$g(x)=-x^{2}$

Part 2) Horizontal shift left 3 units

The rule of the translation is

(x,y) ------> (x-3,-y)

f(x) -----> g(x)

The function g(x) will be

$g(x)=(x+3)^{2}$

Part 3) Vertical shift up 7 units

The rule of the translation is

(x,y) ------> (x,y+7)

f(x) -----> g(x)

The function g(x) will be

$g(x)=x^{2}+7$

5 0

3 years ago

Read 2 more answers

If odd natural numbers are less than 7, then they are prime numbers.

guajiro [1.7K]

I think true?? i think! im really not sure but yea

5 0

2 years ago