AB = 8x + 5 BC = 5x - 9 AC = 74<br><br> what does X, Ab, and BC equal?

What's the expanded form to write 16,107,320

Romashka [77]

(10,000,000) + (6,000,000) + (100,000) + (7,000) + (300) + (20)

6 0

3 years ago

A consumer group is interested in estimating the proportion of packages of ground beef sold at a particular store that have an a

storchak [24]

Answer: 267.

Step-by-step explanation:

When there is no prior information for the population proportion, then the formula we use to find the sample size to estimate the confidence interval :

$n=0.25(\dfrac{z^*}{E})^2$ , where z* = Critical z-value and E + amrgin of error.

Let p = proportion of packages of ground beef sold at a particular store that have an actual fat content exceeding the fat content stated on the label.

Since , we have no prior information about p. so we use above formula

with E = 0.06 and critical value for 95% confidence =z* =1.96 [By z-table ] , we get

$n=0.25(\dfrac{1.96}{0.06})^2=0.25(32.6666)^2\\\\=(0.25)(1067.111111)\\\\=266.777777778\approx267$

Hence, the required sample size is 267.

7 0

3 years ago

In the coordinate plane, three vertices of rectangle abcd are a(0, 0, b(0, a, and d(b, 0. what are the coordinates of point c?

serious [3.7K]

So the point would be c)0,0

3 0

4 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

4 years ago

Which expression is equivalent ?<br><br> LOOK AT THE PICTURE

victus00 [196]

Answer:

x^ (5/3) y ^ 1/3

Step-by-step explanation:

Rewriting as exponents

(x^5y) ^ 1/3

We know that a^ b^c = a^(b*c)

x^ (5/3) y ^ 1/3

8 0

4 years ago

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AB = 8x + 5 BC = 5x - 9 AC = 74 what does X, Ab, and BC equal?