Decide if the following statement is valid or

Write the linear equation in slope intercept form that passes through the given slopes (1,3) and (-3,-5)

Dimas [21]

Answer:

y = -3x

Step-by-step explanation:

The standard expression of equation of a line in slope-intercept form is expressed as;

y = mx+c

m is the slope

c is the intercept

Given

slope m = -3

Point (x, y) = (-1, 3)

x = -1 and y = 3

Get the intercept

To get the intercept c, we will substitute the given values into the equation above to have;

y = mx+c

3 = -3(-1)+c

3 = 3 + c

c = 3-3

c = 0

Substitute m = -3 and c = 0 into the equation above;

y = mx+c

y = -3x+0

y = -3x

Hence the required linear equation is y = -3x

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Delta furnishings received an invoice dated may 10 for a shipment of goods received june 21. The invoice was for a $8400.00 less

djyliett [7]

Answer:

The right solution is:

(a) $1,940

(b) $2,813

Step-by-step explanation:

Given:

Invoice amount

= $8,400

Discount 1

= 33.33%

Discount 2

= 12.5%

So,

The net payable amount will be:

= $8400\times [1 - \frac{100}{300} ]\times [1 - \frac{12.5}{100} ]$

= $8400\times \frac{2}{3}\times 0.875$

= $4900$ ($)

Now,

(a)

Amount paid will be:

= $2000\times 0.97$

= $1940$ ($)

Balance still to be paid will be:

= $4900-2000$

= $2900$ ($)

(b)

Amount paid will be:

= $(4900- 2000)\times 0.97$

= $2900\times 0.97$

= $2813$ ($)

Balance still to be paid will be:

= $2000$ ($)

Thus the above solution is the appropriate one.

6 0

3 years ago

Find the equation of a circle with centre at (3, –2) and passing through the point

Llana [10]

Answer:

beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans beans

6 0

3 years ago

The perimeter of the triangle.

mel-nik [20]

In order to find the perimeter, all we have to do is add all three of the sides:
x^2 + 3 + 2x + 5 + 2x = 29
x^2 + 4x + 8 = 29

Subtract 29 from both sides:
x^2 + 4x - 21 = 0

Factor:
(x + 7)(x - 3) = 0
{-7, 3}

3 can be the only possible length, since a length can't be a negative number.

2x = 6
2x + 5 = 11
x^2 + 3 = 12
-T.B.

4 0

4 years ago

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 83 type K batteries and a sample of 77

agasfer [191]

Answer:

1. Null Hypothesis, $H_0$ : $\mu_1 = \mu_2$ {mean voltage for these two types of

batteries is same}

Alternate Hypothesis, $H_1$ : $\mu_1\neq \mu_2$ {mean voltage for these two types of

batteries is different]

2. Test Statistics value = -5.06

4. Decision for the hypothesis test is that we will reject null hypothesis.

Step-by-step explanation:

We are given that an engineer is comparing voltages for two types of batteries (K and Q).

where, $\mu_1$ = true mean voltage for type K batteries.

$\mu_2$ = true mean voltage for type Q batteries.

So, Null Hypothesis, $H_0$ : $\mu_1 = \mu_2$ {mean voltage for these two types of

batteries is same}

Alternate Hypothesis, $H_1$ : $\mu_1\neq \mu_2$ {mean voltage for these two types of

batteries is different]

The test statistics we use here will be :

$\frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ follows $t_n__1+n_2-2$

where, $X_1bar$ = 9.29 and $X_2bar$ = 9.65

$s_1$ = 0.374 and $s_2$ = 0.518

$n_1$ = 83 and $n_2$ = 77

$s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(83-1)0.374^{2}+(77-1)0.518^{2} }{83+77-2} }$ = 0.45 Here, we use t test statistics because we know nothing about population standard deviations.

Test statistics = $\frac{(9.29-9.65) - 0 }{0.45\sqrt{\frac{1}{83}+\frac{1}{77} } }$ follows $t_1_5_8$

= -5.06

At 0.05 or 5% level of significance t table gives a critical value between (-1.98,-1.96) to (1.98,1.96) at 158 degree of freedom. Since our test statistics is less than the critical table value of t as -5.06 < (-1.98,-1.96) so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that mean voltage for these two types of batteries is different.

7 0

4 years ago