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2 years ago

If you bought 40 cookies at 1.25 and 10 at $1.75 what is your average buy price?

Mathematics

2 answers:

vova2212 [387]2 years ago

6 0

Answer: 67.5

Step-by-step explanation:

i did the math and it took me while but i got the answer i hope this helps

rusak2 [61]2 years ago

4 0

I’m not sure about this but I think the answer is this

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What is each expression written using each base only once? (-4)^-6*(-4)^7

andreyandreev [35.5K]

(-4)^-6*(-4)^7 = (-4)^1 = -4

answer a. -4

6 0

3 years ago

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Sam was given a bag of 100 red and blue marbles. He believes

Rashid [163]

Answer:

(D) There are probably more blue marbles than red marbles in the bag.

Step-by-step explanation:

There are a total of 100 marbles in the bag.

In the experiment of 50 trials, Sam had the following outcome:

Blue=35

Red=15

Relative Frequency of Blue Marbles =35/50=0.7

Relative Frequency of Red Marbles =15/50=0.3

Since the relative frequency of blue marbles is greater than the relative frequency of red marbles, <u>there are probably more blue marbles than red marbles in the bag.</u>

The correct option is D.

6 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

Two events, A and B, are independent of each other. P(A) = 1/6 and P(A and B) = 1/8 What is P(B) written as a decimal? Round to

Leona [35]

P(B) = 0.75.

For independent events, P(A and B) = P(A)*P(B). This gives us

1/8 = 1/6(x)

Divide both sides by 1/6:
1/8 ÷ 1/6 = x
1/8 × 6/1 = x
6/8 = x
3/4 = x
0.75 = x

5 0

4 years ago

What does 3+5x6-1=? and ill mark as brainliest

inna [77]

Answer:

Step-by-step explanation:

3+30-1

33-1

6 0

3 years ago

Read 2 more answers