I will give u brainliest!

Factor 2k^2 - k - 3 A: (2k-3)(k+1) B: (k-1)(2k+3) C: (2k-1)(k+3) D: (k-3)(2k+1)

Art [367]

Answer:

A

Step-by-step explanation:

Given

2k² - k - 3

Consider the factors of the product of the k² term and the constant term which sum to give the coefficient of the k- term.

product = 2 × - 3 = - 6 and sum = - 1

The factors are + 2 and - 3

Use these factors to split the k- term

2k² + 2k - 3k - 3 ( factor the first/second and third/fourth terms )

= 2k(k + 1) - 3(k + 1) ← factor out (k + 1) from each term

= (2k - 3)(k + 1) → A

4 0

3 years ago

A patient has 0.0000075 gram of iron in 1 liter of blood. The normal level is between 6* 10^-7 grams and 1.6*10^-5 grams. Is the

Vinil7 [7]

0.0000075 = 7.5 x 10^6 --- Patients Blood

and it has to be between 0.0000006 and 0.000016

so Yes it is normal

4 0

3 years ago

PLS HELP ILL GIVE U POINTS :)

denpristay [2]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su

Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let $\mu_1$ = mean sales per market in Region 1.

$\mu_2$ = mean sales per market in Region 2.

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, $H_A$ : > $\mu_1-\mu_2\neq$ 0 {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about population standard deviations;

T.S. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean sales in Region 1 = 84

$\bar X_2$ = sample mean sales in Region 2 = 78.3

$s_1$ = sample standard deviation of sales in Region 1 = 6.6

$s_2$ = sample standard deviation of sales in Region 2 = 8.5

$n_1$ = sample of supermarkets from Region 1 = 12

$n_2$ = sample of supermarkets from Region 2 = 17

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times 8.5^{2} }{12+17-2} }$ = 7.782

So, the test statistics = $\frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }$ ~ $t_2_7$

= 1.943

The value of t-test statistics is 1.943.

Now, at a 0.02 level of significance, the t table gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

6 0

3 years ago

The perimeters of the square and the triangle shown below are equal. What is the value of x?

Rudiy27

Perimeter is the sum of all the sides. So we can set up an equation:

$\sf 3x+1+3x+1+3x+1+3x+1=2x+8+x+8+4x+2$

Now solve for 'x', combine like terms:

When it comes to terms with variables it's just like normal addition but we keep the variable:

$\sf 3x+3x+3x+3x=12x$
$\sf 2x+x+4x=7x$

So we have:

$\sf 12x+1+1+1+1=7x+8+8+2$

Add:

$\sf 12x+4=7x+18$

Subtract 7x to both sides:

$\sf 5x+4=18$

Subtract 4 to both sides:

$\sf 5x=14$

Divide 5 to both sides:

$\boxed{\sf x=2.8}$

8 0

3 years ago