Ask question

Ask question

All categories

Ira Lisetskai [31]

3 years ago

5

A total of 75 cookies and cakes were donated for a bake sale to raise money for the football team. There were four times as many

cookies donated as cakes. How many cookies and cakes were donated

2 answers:

MArishka [77]3 years ago

6 0

Answer:

60

Step-by-step explanation:

15*4 = 60

15+60= 75

Serggg [28]3 years ago

6 0

Answer:

There were 15 cakes and 60 cookies. Hope I helped.

Step-by-step explanation:

75-15=60

15x4=60

You might be interested in

consider the following scenario: Assume daily protein intake values in a population of athletes are known to have a standard dev

Kipish [7]

Answer:

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{58.6}{\sqrt{267}} = 7.03$

The lower end of the interval is the sample mean subtracted by M. So it is 77 - 7.03 = 69.97g.

The upper end of the interval is the sample mean added to M. So it is 77 + 7.03 = 84.03g.

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.

5 0

3 years ago

The cities of fayetteville, raleigh and durham form a triangle. they are building a new

Sati [7]

Answer:

it should be built at the centre

6 0

3 years ago

6x-14<-14 or 3x+10>13

kow [346]

Step-by-step explanation:

not sure the question but I can solve the equations

6x-14<-14

6x<0

×<0

3x+10>13

3x>3

×>1

7 0

3 years ago

I need help on these questions please divison with fractions

erica [24]

Charity will have 16 bags

3 0

3 years ago

Read 2 more answers

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