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3 years ago

What is the constant proportionality, in dollars per granola bar

Mathematics

1 answer:

ikadub [295]3 years ago

5 0

Answer:

4 granola bars is $9.00 12 granola bars is $27.00 24 granola bars is $54.00

Step-by-step explanation:

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A cab company charges a $6.00 fee to pick up a fare plus $0.20 per mile. If you pay the cab driver $9.40, how many miles did you

m_a_m_a [10]

Answer is 6.00+0.20-9.40

3.2

6.00-0.20-9.40

3.2

4 0

1 year ago

Please show work! Thank you

xz_007 [3.2K]

A) The greatest rectangular area will be the area of a square 10 m on each side, 100 m^2.

b) The new dimensions will be 11 m × 11 m.
.. The new area will be (11 m)^2 = 121 m^2.

c) The area was increased by 121 m^2 -100 m^2 = 21 m^2, or 21%.

d) Yes, and no.
.. If you increase the dimensions by 10%, the area will increase by 21%.
.. (40 m)^2 = 1600 m^2
.. (44 m)^2 = 1936 m^2 = 1.21*(1600 m^2), an increase of 21% over the original.

.. If you increase the dimensions by 1 unit, the area will increase by (2x+1) square units, where x is the side of the original. For x≠10, this is not 21 square units.
.. (41 m)^2 = 1681 m^2 = 1600 m^2 +(2*40 +1) m^2 = 1600 m^2 +81 m^2

4 0

3 years ago

You buy a six pack of Gatorade for $9.00. What is the unit price or the price per bottle?

Viefleur [7K]

Answer:

Price per bottle is 1.5 or $1.50

Step-by-step explanation:

To get price per unit, you just divide the amount of money spent by the items purchased. 9/6 = 1.5

4 0

3 years ago

How do I do this help

tatuchka [14]

6÷40.5= $6.75 per hour
8÷54= $6.75 per hour
y= 6.75x

5 0

3 years ago

The amount of time that people spend at Grover Hot Springs is normally distributed with a mean of 73 minutes and a standard devi

Vesnalui [34]

Answer:

(a) $X\sim N(\mu = 73, \sigma = 16)$

(b) 0.7910

(d) 0.6464

Step-by-step explanation:

Let X = amount of time that people spend at Grover Hot Springs.

The random variable X is normally distributed with a mean of 73 minutes and a standard deviation of 16 minutes.

(a)

The distribution of the random variable X is:

$X\sim N(\mu = 73, \sigma = 16)$

(b)

Compute the probability that a randomly selected person at the hot springs stays longer than 60 minutes as follows:

$P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-73}{16})\\=P(Z>-0.8125)\\=P(Z$

*Use a z-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays longer than an hour is 0.7910.

(c)

Compute the probability that a randomly selected person at the hot springs stays less than 45 minutes as follows:

$P(X$

*Use a z-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays less than 45 minutes is 0.0401.

(d)

Compute the probability that a randomly person spends between 60 and 90 minutes at the hot springs as follows:

$P(60$

*Use a z-table for the probability.

Thus, the probability that a randomly person spends between 60 and 90 minutes at the hot springs is 0.6464

6 0

3 years ago