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fomenos
3 years ago
10

While at the beach Daniel buys lunch for his family from a food stand he purchased one hotdog for $2.50 and three hamburgers if

you spent $13 total right and solve an equation to find H the amount of each hamburgers cost
Mathematics
1 answer:
Alex17521 [72]3 years ago
7 0

The cost of each hamburger is $ 3.5

<em><u>Solution:</u></em>

Given that, Daniel purchased one hotdog for $2.50 and three hamburgers

He spent $ 13 in total

Let "H" be the cost of each hamburger

From given,

one hotdog for $2.50 + three hamburgers each for "H" = $ 13

Thus we get,

1 \times 2.50 + 3 \times H = 13\\\\2.50 + 3H = 13\\\\\text{Solve the above equation for H}\\\\3H = 13 - 2.50\\\\3H = 10.5\\\\\text{Divide both sides of equation by 3}\\\\H = 3.5

Thus cost of each hamburger is $ 3.5

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Step-by-step explanation:

Given that a store receives shipments of peaches.  They want to sort them by weight.

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How many cups of flour are needed if we increase the recipe by 50%?
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What should I buy? A study conducted by a research group in a recent year reported that of cell phone owners used their phones i
Llana [10]

Answer:

The probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7886.

<em />

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of 14 cell phone owners is studied. Round the answers to at least four decimal places. What is the probability that seven or more of them used their phones for guidance on purchasing decisions? </em>

We can model this as a binomial random variable, with p=0.57 and n=14.

P(x=k)=\dbinom{n}{k} p^{k}q^{n-k}

a) We have to calculate the probability that seven or more of them used their phones for guidance on purchasing decisions:

P(x\geq7)=\sum_{k=7}^{14}P(x=k)\\\\\\

P(x=7)=\dbinom{14}{7} p^{7}q^{7}=3432*0.0195*0.0027=0.1824\\\\\\P(x=8) = \dbinom{14}{8} p^{8}q^{6}=3003*0.0111*0.0063=0.2115\\\\\\P(x=9) = \dbinom{14}{9} p^{9}q^{5}=2002*0.0064*0.0147=0.1869\\\\\\P(x=10) = \dbinom{14}{10} p^{10}q^{4}=1001*0.0036*0.0342=0.1239\\\\\\P(x=11) = \dbinom{14}{11} p^{11}q^{3}=364*0.0021*0.0795=0.0597\\\\\\P(x=12) = \dbinom{14}{12} p^{12}q^{2}=91*0.0012*0.1849=0.0198\\\\\\P(x=13) = \dbinom{14}{13} p^{13}q^{1}=14*0.0007*0.43=0.004\\\\\\

P(x=14) = \dbinom{14}{14} p^{14}q^{0}=1*0.0004*1=0.0004\\\\\\

P(x\geq7)=0.1824+0.2115+0.1869+0.1239+0.0597+0.0198+0.004+0.0004\\\\P(x\geq7)=0.7886

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