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sergiy2304 [10]
2 years ago
13

Allie has $20 to spend to take some friends bowling. It costs $4 per person to rent shoes. The bowling alley is running a specia

l where each game costs $1, no matter how many players. If Allie and her friends want to play three games and each person has to rent shoes, how many friends can Allie bring? *
Mathematics
2 answers:
lukranit [14]2 years ago
6 0

Answer:

4 as a fraction 4 1/17

Step-by-step explanation:

20 - 3 = 17

17 divid 4

= 4

forsale [732]2 years ago
4 0

Answer:

Step-by-step explanation: she can have four people go and have the last 4 dollers can go in to her buying the lanes

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Which expression is equivalent to -16 - 7?
lakkis [162]

Step-by-step explanation:

B. -16 + (-7) ...this is the same as -16-7

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2 years ago
Corinne made a deposit of $6,000 into an account with a simple interest rate of 5%. How much interest will be in the account aft
oksian1 [2.3K]

Answer:

$8,864.73

Step-by-step explanation:

Base amount: $6,000.00

Interest Rate: 5% (yearly)

Effective Annual Rate: 5%

Calculation period: 8 years

8 0
3 years ago
CAN ANYONE HELP WITH THESE? BRAINLIEST ANS
Sonbull [250]
The answer to the first one is a

3 0
2 years ago
Two cars leave Phoenix and travel along roads 90 degrees apart. If Car 1 leaves 30 minutes earlier than Car 2 and averages 42 mp
romanna [79]

Answer:

C. 210 miles

Step-by-step explanation:

We have been given that two cars leave Phoenix and travel along roads 90 degrees apart. Car 1 leaves 30 minutes earlier than Car 2 and averages 42 mph.

We will use distance formula and Pythagoras theorem to solve our given problem.

\text{Distance}=\text{Speed}\times \text{Time}

\text{Distance covered by Car 1 in 3.5 hours}=\frac{42\text{ Miles}}{\text{Hour}}\times \text{3.5 hour}

\text{Distance covered by Car 1 in 3.5 hours}=147\text{ Miles}.

Since car 1 leaves 30 minutes before car 2, so car 2 will travel for only 3 hours when car 1 will travel for 3.5 hours.

\text{Distance covered by Car 2 in 3 hours}=\frac{50\text{ Miles}}{\text{Hour}}\times \text{3 hour}

\text{Distance covered by Car 2 in 3 hours}=150\text{ Miles}

Since both car travel along roads 90 degree apart, therefore, the distance between both cars after Car 1 has traveled 3.5 hours would be hypotenuse with legs 147 and 150.

\text{Distance between both cars}=\sqrt{147^2+150^2}

\text{Distance between both cars}=\sqrt{21609+22500}

\text{Distance between both cars}=\sqrt{44109}

\text{Distance between both cars}=210.021427\approx 210

Therefore, the both cars will be 210 miles apart and option C is the correct choice.

3 0
3 years ago
Two surveys were independently conducted to estimate a population mean, μ. denote the estimates and their standard errors by x1
lyudmila [28]

Hi there what you need is lagrange multipliers for constrained minimisation. It works like this,

V(X)=α2σ2X¯1+β2\sigma2X¯2

Now we want to minimise this subject to α+β=1 or α−β−1=0.

We proceed by writing a function of alpha and beta (the paramters you want to change to minimse the variance of X, but we also introduce another parameter that multiplies the sum to zero constraint. Thus we want to minimise

f(α,β,λ)=α2σ2X¯1+β2σ2X¯2+λ(\alpha−β−1).

We partially differentiate this function w.r.t each parameter and set each partial derivative equal to zero. This gives;

∂f∂α=2ασ2X¯1+λ=0

∂f∂β=2βσ2X¯2+λ=0

∂f∂λ=α+β−1=0

Setting the first two partial derivatives equal we get

α=βσ2X¯2σ2X¯1

Substituting 1−α into this expression for beta and re-arranging for alpha gives the result for alpha. Repeating the same steps but isolating beta gives the beta result.

Lagrange multipliers and constrained minimisation crop up often in stats problems. I hope this helps!And gosh that was a lot to type!xd

3 0
3 years ago
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