write a ratio for the following description: for every 6 cups of flour in bread recipe, there are 2 cups of milk

9.7 x 10^3 in scientific notation

harina [27]

The answer to the question is 9,700.

8 0

3 years ago

Factor completely.<br> 2x^4 + 4x^3 - 30x^2 =

den301095 [7]

Answer:

$2x^2(x-3)(x+5)$

Step-by-step explanation:

The first thing that you can do is to factor out an x^2 from the expression.

x^2(2x^2+4x-30)

The next step is to factor out a 2.

2x^2(x^2+2x-15)

Finally, you can factor the quadratic inside.

2x^2(x-3)(x+5)

And now you're finished! Hope this helps!

6 0

4 years ago

Read 2 more answers

Which linear inequality is represented by the graph y<1/3x-1.3

balu736 [363]

The answer is A. y<1/3-1.3

3 0

4 years ago

For a treasure hunt game. 300 balls are hidden, 40% of them are yellow and the rest are white. sally's team finds 20% of the yel

Verizon [17]

24 yellow balls and 144 white balls are found

3 0

4 years ago

Read 2 more answers

A special-interest group has conducted a survey concerning a ban on hand guns. Note: A rifle is a gun, but it is not a hand gun.

ivanzaharov [21]

Answer:

a)109 of the surveyed households only own a hand gun and do not support the ban on hand guns.

b)328 of the surveyed households do not own a gun and support the ban on hand guns

c )112 of the surveyed households do not own a gun and do not support the ban on handguns.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents those who own a hand gun.

-The set B represents those who own a rifle

-The set C represents those who support the ban of handguns.

-D represents those who do not own a gun and do not support the ban on hand guns.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which $a$ are those that only own a hand gun, $A \cap B$ are those who own both a handgun and a rifle, $A \cap C$ are those who own a hand gun and support the ban on handguns, and $A \cap B \cap C$ are those who own both a hand gun and a rifle, and also support the ban on hand guns.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

In which $b$ are those who only own a rifle, and $B \cap C$ are those who own a rifle and support the ban on handguns.

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

In which $c$ are those who support the ban on hand guns and do not own any handguns.

This diagram has the following subsets:

$a,b,c,D(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

There were 1000 households surveyed, so:

$a + b + c + D + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1000$

We start finding the values from the intersection of three sets.

Solution:

48 own both a hand gun and a rifle and also support the ban on hand guns.

$A \cap B \cap C = 48$

70 own a hand gun and support the ban on hand guns:

$(A \cap C) + (A \cap B \cap C) = 70$

$A \cap C = 70 - 48$

$A \cap C = 22$

206 own a rifle but no hand gun and do not support the ban on hand guns.

$b = 206$ .

136 own both a hand gun and a rifle:

$(A \cap B) + (A \cap B \cap C) = 136$

$A \cap B = 136 - 48$

$A \cap B = 88$

493 supported the ban on hand guns.

$C = 493$

437 own a rifle:

$B = 437$

267 own a hand gun:

$A = 267$

a) How many of the surveyed households only own a hand gun and do not support the ban on hand guns?

This is the value of $a$ , that we can find from the following equation:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

$267 = a + 88 + 22 + 48$

$a = 267 - 158$

$a = 109$

109 of the surveyed households only own a hand gun and do not support the ban on hand guns.

b) How many of the surveyed households do not own a gun and support the ban on hand guns?

This is the value of $c$ , that we can find from the following equation:

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$493 = c + 22 + (B \cap C) + 48$

Here, we also have to find $B \cap C$ , that are those who own a rifle and support the ban on hand guns. We can find this from the following equation:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$437 = 206 + (B \cap C) + 88 + 48$