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aniked [119]
2 years ago
10

Hailey paid \$13$13dollar sign, 13 for 1\dfrac3{7} \text{ kg}1 7 3 ​ kg1, start fraction, 3, divided by, 7, end fraction, start

text, space, k, g, end text of sliced salami. What was the cost per kilogram of salami
Mathematics
2 answers:
il63 [147K]2 years ago
4 0

Answer:

91/10 :)

Step-by-step explanation:

LenKa [72]2 years ago
3 0

Answer:

The cost of per kilogram of salami is = $9.1

Step-by-step explanation:

Given:

Hailey paid $13 for 1\frac{3}{7} kg of sliced salami.

To find cost per kilogram of salami.

Solution:

We will apply unitary method to find per kilogram cost of salami.

If 1\frac{3}{7} kg of salami costs = $13

Then 1 kg of salami will cost in dollars will be = 13\div 1\frac{3}{7}

In order to divide by mixed numbers, we will convert mixed numbers to fractions.

⇒ 13\div \frac{(7\times1)+3}{7}

⇒ 13\div \frac{10}{7}

In order to divide by fractions we will take reciprocal of divisor and then multiply.

⇒ 13\times \frac{7}{10}

⇒ \frac{13\times7}{10}

⇒ \frac{91}{10}

⇒ 9.1

Thus, cost of per kilogram of salami is = $9.1

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A trader paid 1500 for 6 drinking cups. one of the cups got broken. he sold the remaining five, making a profit of 10percent cal
tankabanditka [31]

Answer:

Cost price of each cup = 1500/6 = 250

Selling price of each cup = 1650/5 = 330

Step-by-step explanation:

Cost price of cups = 1500

Profit = 10%

Selling price of cups = ?

Profit and loss percentage are calculated based on cost price and the formula to calculate profit is mentioned as under:

Profit percentage (%) = Profit x 100/Cost price

10 = Profit x 100/1500

⇒ Profit = 10 x 1500/100

⇒ Profit = 150

The formula for finding selling price is mentioned as under:

Selling Price = Profit + Cost Price.

Selling Price = 150 + 1500

⇒ Selling Price = 1650.

Since one of the six cups was broken that means the seller earned profit based on only 5 cups, it implies that selling price of each cup will be 1650/5 = 330.

6 0
2 years ago
Yohanne and Mikaela completed a 1-km race. Yohanne finished the race 1 minute after Mikaela. If the speed of Mikaela is 50 meter
lilavasa [31]

Answer: 5 min

Step-by-step explanation:

Given

Length of the race track l=1\ km\ or\ 1000\ m

Suppose the speed of Yohanne is v_y=x\ m/min.

then  Mikaela speed is v_m=50+x

the difference in the finishing time is 1 minute i.e.

\Rightarrow \dfrac{1000}{x}-\dfrac{1000}{50+x}=1\\\\\Rightarrow \dfrac{50+x-x}{x(50+x)}=\dfrac{1}{1000}\\\\\Rightarrow \dfrac{50}{x(50+x)}=\dfrac{1}{1000}\\\\\Rightarrow x^2+50x-50,000=0\\\\\Rightarrow x^2+250x-200x-50,000=0\\\\\Rightarrow (x+250)(x-200)=0\\\\\Rightarrow x=200\ m/min.

Time taken by Yohanne is

\Rightarrow t=\frac{1000}{200}=5\ min

7 0
1 year ago
3.12 Speeding on the I-5, Part I. The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in Ca
marusya05 [52]

Answer:

a) 93.943% = 93.9%

b) 93.528% = 93.5%

c) Speed of the fastest 5% ≥ 80.5 miles/hour

d) 29.46% = 29.5%

Step-by-step explanation:

Mean, xbar = 72.6 miles/hour.

standard deviation, σ = 4.78 miles/hour

For each of the questions, we'll need to normalize the speeds.

a) The standardized score for 80 miles/hour is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ = (80 - 72.6)/4.78 = 1.55

To determine the probability of a car having speed less than 80 miles/hour, P(x < 80) = P(z < 1.55)

We'll use data from the normal probability table for these probabilities

P(x < 80) = P(z < 1.55) = 1 - P(z ≥ 1.55) = 1 - P(z ≤ -1.55) = 1 - 0.06057 = 0.93943

b) percent of passenger vehicles travel between 60 and 80 miles/hour.

60 miles/hour standardized = (60 - 72.6)/4.78 = -2.64

We'll use data from the normal probability table for these probabilities

P(60 < x < 80) = P(-2.64 < z < 1.55) = P(z ≤ 1.55) - P(z ≤ -2.64) = 0.93943 - 0.00415 = 0.93528

c) How fast to do the fastest 5% of passenger vehicles travel?

We'll use data from the normal probability table for these probabilities

Top 5% corresponds to a z-score of 1.65. P(z ≥ 1.65) = 0.95053

1.65 = (x - 72.6)/4.78

x = 80.487 miles/hour = 80.5 miles/hour.

d) The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.

70 miles/hour, standardized = (70 - 72.6)/4.78 = 0.54

P(x > 70) = P(z > 0.54) = 1 - P(z ≤ 0.54) = 1 - 0.7054 = 0.2946.

Hope this helps!!!!

6 0
2 years ago
The population of a town is growing uninhibited at a rate of 1% per year. How long would it take the population to double? Round
lianna [129]
For this question the answer is d
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Consider the following regression model: Humidity = β0 + β1Temperature + β2Spring + β3Summer + β4Fall + β5Rain + ε, where the du
FinnZ [79.3K]

Answer:

The regression equation for the winter rainy days is "Humidity = (β0 + β5) + β1Temperature".

Step-by-step explanation:

Given:

Humidity = β0 + β1Temperature + β2Spring + β3Summer + β4Fall + β5Rain + ε ...........(1)

Since there can be only one of spring, summer,fall, and winter at a point in time or in a season, we will have the following when there are winter rainy days:

Spring = 0

Summer = 0

Fall = 0

Rain = 1

Substituting all the relevant values into equation (1) and equating ε also to 0, a reduced form of equation (1) can be obtained as follows:

Humidity = β0 + β1Temperature + (β2 * 0) + (β3 * 0) + (β4 * 0) + (β5 * 1) + 0

Humidity = β0 + β1Temperature + 0 + 0 + 0 + β5 + 0

Humidity = (β0 + β5) + β1Temperature

Therefore, the regression equation for the winter rainy days is "Humidity = (β0 + β5) + β1Temperature".

3 0
2 years ago
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