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2 years ago

What is the area of the composite shape? A.14 B.24 C.36 D.68

Mathematics

1 answer:

lyudmila [28]2 years ago

6 0

Answer:

Step-by-step explanation:

The top part is 4.5 inches and teh bottom is 27 so it is 36

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A scientist mixes water (containing no salt) with a solution that contains 35% salt. She wants to obtain 245 ounces of a mixture

sesenic [268]

Answer:

140 ounces of water and 105 ounces of 35% salt solution.

Explanation:

Let $x$ be the ounces of water that need to be added and $y$ be the ounces of 35% solution that need to to be added.

We know that the total of ounces of water and of 35% solution must be 245 ounces:

$x+y=245$

35% of $x$ is the salt, and since no salt is added with addition of $y$ liters of water, this must be equal to the amount of salt in the final 15% solution:

$0.35x=0.15*245$

we solve for $x$ and get:

$x=105$

We put this value into $x+y=245$ and solve for $y$ to get:

$y=140$

Thus we have 140 ounces of water and 105 ounces of 35% solution.

8 0

3 years ago

Nancy had 16 chocolate candies in a bag. Her mother put a handful of candies in the bag. When Nancy counts her chocolate candies

ryzh [129]

The equation should look something like:

x+16 = 32

16+x = 32

5 0

3 years ago

Read 2 more answers

If f(x) varies directly with x and f(x) = 4 when x = 12, then what is the value of f(x) when x = 60?

ANTONII [103]

ANSWER

$f(60)=3\times60=180$

EXPLANATION
We were given that
$f(x)$
varies directly as
$x.$

We can write this mathematically as,

$f(x) \propto \: x.$

This implies that,

$f(x) = kx$
where k is the constant of variation.

$f(4)=4k$

This implies that,

$4k=12$

$k = 3$

The equation becomes

$f(x)=3x$

When $x=60$

$f(60)=3\times60=180$

6 0

3 years ago

If gas is $2.45 and you pay $ 12.00 how many gallons of gas do you get ?

attashe74 [19]

Gas is 2.45 per gallon. So if we pay $12 we get 12÷2.45 gallons:

12 ÷ 2.45 ≈ 4.90 gallons.

You get roughly 4.90 gallons.

6 0

2 years ago

Read 2 more answers

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$