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3 years ago

5

The probability distribution below shows the number of kittens in a litter, Find the probability that a cat will have at most 4

kittens * 7 points

2 answers:

jonny [76]3 years ago

6 0

Answer:

Step-by-step explanation:

.28

zalisa [80]3 years ago

4 0

Plz help I’m getting timed

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The price of a car was increased from $3,600 to $4,500. What was the percent of the increase?

Nataly [62]

Answer:

25%

Step-by-step explanation:

.25(3,600)=900

3,600+900=4,500

6 0

3 years ago

Read 2 more answers

Please help me out :( I appreciate it

lilavasa [31]

The answer is A because The median for the wolverines, 80, is more than the median for the panthers, 70.

The “median” is the “Middle” value in the list of numbers.

3 0

3 years ago

.75x-20= 620 what does x equal rounded to the nearest 100?

vodomira [7]

To solve this, we’ll have to use inverse operations.

Since addition is the inverse to subtraction, we’ll add 20 to 620.

20 + 620 = 640

From there, since division is the inverse of multiplication, we’ll divide .75 by 640.

640/.75 = 853.33333 repeating. We can round this to 853.33, to round to the nearest hundredth.

Therefore, 853.33 would be your answer! You can check this by solving the equation forwards (853.33 x .75 = ~640, 640 - 20 = 620.)

Hope this helps! :)

4 0

3 years ago

A cake is removed from a 310°F oven and placed on a cooling rack in a 72°F room. After 30 minutes the cake's temperature is 220°

Fynjy0 [20]

Answer:

The time is 135 min.

Step-by-step explanation:

For this situation we are going to use Newton's Law of Cooling.

Newton’s Law of Cooling states that the rate of temperature of the body is proportional to the difference between the temperature of the body and that of the surrounding medium and is given by

$T(t)=C+(T_0-C)e^{kt}$

where,

C = surrounding temp

$T(t)$ = temp at any given time

t = time

$T_0$ = initial temp of the heated object

k = constant

From the information given we know that:

Initial temp of the cake is 310 °F.
The surrounding temp is 72 °F.
After 30 minutes the cake's temperature is 220 °F.

We want to find the time, in minutes, since the cake's removal from the oven, at which its temperature will be 100°F.

To do this, first, we need to find the value of k.

Using the information given,

$220=72+(310-72)e^{k\cdot 30}\\\\72+238e^{k30}=220\\\\238e^{k30}=148\\\\e^{k30}=\frac{74}{119}\\\\\ln \left(e^{k\cdot \:30}\right)=\ln \left(\frac{74}{119}\right)\\\\k\cdot \:30=\ln \left(\frac{74}{119}\right)\\\\k=\frac{\ln \left(\frac{74}{119}\right)}{30}$

$T(t)=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}$

Next, we find the time at which the cake's temperature will be 100°F.

$100=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}\\72+238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=100\\238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=28\\e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=\frac{2}{17}\\\ln \left(e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}\right)=\ln \left(\frac{2}{17}\right)\\\frac{\ln \left(\frac{74}{119}\right)}{30}t=\ln \left(\frac{2}{17}\right)\\t=\frac{30\ln \left(\frac{2}{17}\right)}{\ln \left(\frac{74}{119}\right)}\approx 135.1$

4 0

3 years ago

What values of a and b make the equation true?

mestny [16]

Answer:

3

Step-by-step explanation:

$\sqrt{648} =\sqrt{2 * 2 * 2 * 3 * 3 * 3 * 3} \\\\=\sqrt{2^{3}*3^{4}}$

a = 3 & b =4

6 0

3 years ago