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VLD [36.1K]
3 years ago
15

At a store, Lauren selected a watermelon that weighed 38.4 ounces.

Mathematics
1 answer:
Mrrafil [7]3 years ago
5 0

Answer:

$3.60

Step-by-step explanation:

1.50 ÷ 16 = 0.09375

there are 38.4 oz in laurens watermelon, so multiply the cost per ounce by the number of ounces to find the answer

0.09375 × 30.4 = 3.6

You might be interested in
Salaries of entry-level computer engineers have Normal distribution with unknown mean and variance. Three randomly selected comp
avanturin [10]

Answer:

H0 : μ = 60000

H1 : μ ≠ 60000

Test statistic = 3.464

Step-by-step explanation:

Given :

Sample mean salary, xbar = 80000

Sample standard deviation, s = 10000

Population mean salary , μ = 60000

Sample size, n = 3

Hypothesis :

H0 : μ = 60000

H1 : μ ≠ 60000

The test statistic :

T = (xbar - μ) ÷ (s/√(n))

T = (80000 - 60000) ÷ (10000/√(3))

T = 20000 / 5773.5026

T = 3.464

The Decison region :

If Tstatistic >Tcritical

Tcritical at 10%, df = 2 ; two - tailed = 2.9199

Tstatistic > Tcritical ; He

5 0
3 years ago
The weekly salaries of six employees at McDonalds are $140, $220, $90, $180, $140, $200. For these six salaries, determine: (a)
lys-0071 [83]

Answer:

a) Mean = 161.666667

b)Mode = 140

c) Median =160

d) Range =130

Step-by-step explanation:

Given that the weekly salaries of six employees =$140, $220, $90, $180, $140, $200

Mean =Total number of salaries/ Number of salaries

=$140 +$220 +$90  +$180 + $140 +$200  /6=970/6= 161.666667

Mode = The number with the highest frequency ie the number that occurs most in the listed salaries

$140, $220, $90, $180, $140, $200

Therefore 140 is the mode

Median = The number that occurs in the middle, Now that we have a list of even Numbers, our median would be the sum of the two middle numbers divided by 2 after arranging in ascending order fro  least to greatest

$90, $140,$140, $180, $200,$220

Median =$140+ $180  / 2 =  160

Range = The difference between the highest and lowest number in the sample

$220-$90=130

7 0
3 years ago
Seattle-Pipes Co. produces pipes to be supplied to a Seattle utility company. The requirement of the utility company is that the
Goshia [24]

Answer:

Step-by-step explanation:

Hello!

The study variable is

X: Pipe length.

It is known that this variable has a normal distribution and that the distribution parameter varies depending on the process used to manufacture the pipes.

Process A: μ= 200cm δ= 0.5cm

Process B: μ=201cm δ= 1cm

Process C: μ=202cm δ= 1.5cm

Pipes with a length of 200cm or more will be accepted by the utility company (X≥200), but pipes with less than 200cm length will be rejected (X<200)

a. Using Process C, you need to calculate the probability that the pipe will be rejected, symbolically:

P(X<200)

Using the distribution data of process C you have to standardize the value:

P(Z<(200-202)/1.5)= P(Z<-1.33)= 0.09176

b. The requirements change, accepting any pipe between 199 and 202, you have to calculate the probabilities of the pipes being between those lengths using the three process:

Process A:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-200)/0.5)) - P(Z≤(199-200)/0.5))

P(Z≤4) - P(Z≤-2) = 1 - 0.02275 = 0.97725

The probability of the pipe being rejected is 0.02275

Process B:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-201)/1)) - P(Z≤(199-201)/1))

P(Z≤1) - P(Z≤-2) = 0.84134 - 0.02275 = 0.81859

The probability of the pipe being rejected is 1-0.81859= 0.18141

Process C:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-202)/1.5)) - P(Z≤(199-202)/1.5))

P(Z≤0) - P(Z≤-2) = 0.5 - 0.02275 = 0.47725

The probability of the pipe being rejected is 1-0.47725= 0.52275

The pipes manufactured using process A has fewer chances of being rejected.

c.

Process A costs $140

P(X≥200)= 1 - P(X<200)= 1 - P(Z<0)= 1 - 0.5= 0.5

Process B costs $160

P(X≥200)= 1 - P(X<200)= 1 - P(Z<-1)= 1 - 0.15866= 0.84134

Process C costs $177

P(X≥200)= 1 - P(X<200)= 1 - P(Z<-1.33)= 1 - 0.09176= 0.90824

If they where to make 100 pipes:

Using process A: 100*0.5= 50 pipes will be accepted, so they'll win 50*($200-$140)= $3000

Using process B: 100*0.84134= 84.134≅ 84 pipes will beaccepted, so they'll win 84*($200-$160)= $3360

Using the process C: 100*0.90824= 80.824≅ 90 pipes will be accepted, so they'll win 90*($200-$177)= $2070

As you can see, using process B will maximize the profits.

I hope it helps!

6 0
4 years ago
6. A group of students was given the following to simplify: 3x+7. Brian believes that this expression simplifies to 10x because
PIT_PIT [208]
Marcos is correct because 3x is not a like term to 7, since you cannot add them it’s already in its most simplified form and stays as 3x + 7.
7 0
3 years ago
ohn is interested in purchasing a multi-office building containing five offices. The current owner provides the following probab
Galina-37 [17]

Answer:

$28,875

Step-by-step explanation:

The given table is:

\left|\begin{array}{c|cccccc}$Number of Lease Offices&0&1&2&3&4&5\\$Probability&5/32&3/16&3/32&1/4&9/32&1/32\end{array}\right|

The expected probability is derived using the formula: \sum_{i=0}^{5}x_iP(x_i)

Therefore, for the given distribution,

Expected Probability

=(0*\frac{5}{32})+ (1*\frac{3}{16})+ (2*\frac{3}{32})+ (3*\frac{1}{4})+ (4*\frac{9}{32})+ (5*\frac{1}{32})\\=2.40625

If each yearly lease is $12,000

The Expected Yearly Lease for the Whole Building=2.40625 X 12000

=$28,875

4 0
3 years ago
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