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victus00 [196]
3 years ago
5

Can someone solve my homework questions please ?

Mathematics
1 answer:
gulaghasi [49]3 years ago
4 0
A is not congruent
B is not congruent
C is congruent
D is not congruent
E is congruent
Is congruent
You might be interested in
Solve: the quantity 2x minus 20 divided by 3 = 2x
ira [324]

Answer:

x=-5

Step-by-step explanation:

\dfrac{2x-20}{3}=2x

Multiply both sides by 3:

2x-20=6x

Subtract 2x from both sides:

-20=4x

Divide both sides by 4:

x=-5

Hope this helps!

6 0
3 years ago
Read 2 more answers
A watering can dispenses water at the rate of 0.25 gallon per minute. The original volume of water in the can was 6 gallons. Whi
nydimaria [60]

I believe the answer should be A {(1, 6.0),(2, 5.75),(3, 5.50)}

4 0
3 years ago
Ms. Nina has 10 lbs of 25% sugar syrup. How much water does she need to add to make 10% sugar syrup?
ryzh [129]
She has 10lbs of 25% syrup... so, in the 10lbs, 25% of that is syrup, the rest, namely the 75% remaining is water or other substances.

let's say she adds "x" lbs of water, to get "y" lbs for the 10% mixture.

how much is 25% of 10lbs?  well, (25/100) * 10, or 2.5.

the water has no sugar syrup in it, so is just pure water, so the amoun of syrup in it is 0%, how much is 0% of "x" lbs?  well, (0/100) * x, or 0.00x, which is just 0.

how much is 10% of "y" lbs?  well (10/100) * y, or 0.10y.

whatever "x" and "y" are, we know that 10 + x = y.

we also know that the syrup amount in that is also 2.5 + 0.00x = 0.10y, thus


\bf \begin{array}{lccclll}
&\stackrel{lbs}{syrup}&\stackrel{concentration~\%}{syrup}&\stackrel{concentration}{amount}\\
&------&------&------\\
\textit{25\% syrup}&10&0.25&2.5\\
\textit{pur water}&x&0.00&0.00x\\
------&------&------&------\\
\textit{10\% mixture}&y&0.10&0.10y
\end{array}
\\\\\\
\begin{cases}
10+x=\boxed{y}\\
2.5+0.00x=0.10y\\
----------\\
2.5 = 0.10\left( \boxed{10+x} \right)
\end{cases}
\\\\\\
2.5=1 + 0.10x\implies 1.5=0.10x
\\\\\\
\cfrac{1.5}{0.10}=x\implies 15=x
8 0
3 years ago
Read 2 more answers
We are throwing darts on a disk-shaped board of radius 5. We assume that the proposition of the dart is a uniformly chosen point
Vlad1618 [11]

Answer:

the probability that we hit the bullseye at least 100 times is 0.0113

Step-by-step explanation:

Given the data in the question;

Binomial distribution

We find the probability of hitting the dart on the disk

⇒ Area of small disk / Area of bigger disk

⇒ πR₁² / πR₂²

given that; disk-shaped board of radius R² = 5, disk-shaped bullseye with radius R₁ = 1

so we substitute

⇒ π(1)² / π(5)² = π/π25 = 1/25 = 0.04

Since we have to hit the disk 2000 times, we represent the number of times the smaller disk ( BULLSEYE ) will be hit by X.

so

X ~ Bin( 2000, 0.04 )

n = 2000

p = 0.04

np = 2000 × 0.04 = 80

Using central limit theorem;

X ~ N( np, np( 1 - p ) )

we substitute

X ~ N( 80, 80( 1 - 0.04 ) )

X ~ N( 80, 80( 0.96 ) )

X ~ N( 80, 76.8 )

So, the probability that we hit the bullseye at least 100 times, P( X ≥ 100 ) will be;

we covert to standard normal variable

⇒ P( X ≥  \frac{100-80}{\sqrt{76.8} } )

⇒ P( X ≥ 2.28217 )

From standard normal distribution table

P( X ≥ 2.28217 ) = 0.0113

Therefore, the probability that we hit the bullseye at least 100 times is 0.0113

3 0
3 years ago
Lake Michigan's volume is approximately 1,180 cubic miles and its surface area is approximately 14,332,090 acres. The 2015 water
galben [10]

Answer:

a.)   depth of the lake = (volume of the lake) / area of lake

            = 173,693,585,360,000 / 624,305,840,400  = 278.22 feet

b.) 11 inches  = 0.9167 feet

    Water gained in cubic feet gained by the lake

         = 624,305,840,400 \times  0.9167  

         = 572,282,434,719.5 cubic feet

c.) the percentage change in the lake volume over that year

      = 572,282,434,719.5 cubic feet / 173,693,585,360,000 cubic feet

      =  0.0033

      = 0.33%

Step-by-step explanation:

i) acres to square feet :    1 acre = 43560 square feet

  therefore 14,332,090 acres  = 624,305,840,400 square feet

ii) 1 mile  = 5280 feet

   1 cubic mile  = 5280 \times 5280 \times 5280 = 147,197,952,000 cubic feet

therefore 1180 cubic miles  = 173,693,585,360,000 cubic feet

a.)   depth of the lake = (volume of the lake) / area of lake

            = 173,693,585,360,000 / 624,305,840,400  = 278.22 feet

b.) 11 inches  = 0.9167 feet

    Water gained in cubic feet gained by the lake

    = 624,305,840,400 \times  0.9167   = 572,282,434,719.5 cubic feet

c.) the percentage change in the lake volume over that year

  = 572,282,434,719.5 cubic feet / 173,693,585,360,000 cubic feet

   =  0.0033

  = 0.33%

7 0
3 years ago
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