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labwork [276]
3 years ago
7

There are five puppies in a litter. If the probability that a puppy is male is 0.5, what is the probability that at most one pup

py in the litter is male?
a. 0.15625
b. 0.1875
c. 0.5
d. 0.8125
Mathematics
1 answer:
atroni [7]3 years ago
8 0
I believe C. 0.5 because if the probability that 1 puppy is a male and they are asking what is the probability is that at most one would be male that would seem as the same as saying one male puppy has a 0.5 probability. Hope it help and isn't confusing how I worded it.

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Lillian works 7hours each day for 5 days a week she earn £420 each week how much does she earn per hour
frosja888 [35]

Answer:

£12 per hour

Step-by-step explanation:

7 × 5 = 35 hours

35 h : 420

1 h : x

x = 420×1 / 35

= £12 per hour

8 0
3 years ago
Is 48.4 closest to 49 on a number line
liubo4ka [24]

Answer:

if there are any less than 4.84 then yes it is.

Step-by-step explanation:

please give brainliest and hope this helps

3 0
3 years ago
Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0
2 years ago
Neptunes average distance from the sun is 4.503x10^9 km. Mercurys average distance from the sun is 5.791x10^7 km. About how many
LuckyWell [14K]
Neptune is 77.76 times farther from the Sun than Mercury is. 


6 0
3 years ago
The length of the rectangular garden is 4 meters longer than the width. If the garden area is 60 square
rusak2 [61]

Step-by-step explanation:

Let x be the Width, and x+4 be the Length of the rectangle ; =>

Area = x*(x+4) = 60 ; =>

x^2 +4x - 60 = 0 ; Now solve the Quadratic equation using factoring to find x;

<h2>MARK ME BRAINLIEST!!</h2>
5 0
2 years ago
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