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Svetlanka [38]
3 years ago
8

Larry harvested 4/5 of his crops already he is able to harvest 1/10 of the crops per day how many days will it take for Larry to

harvest all of his crops
Mathematics
1 answer:
Lera25 [3.4K]3 years ago
5 0

Answer:

2 days

Step-by-step explanation:

there is still 1/5 of his crops that need to be harvested. to see how many days that will take divide that amount and see how many 1/10 fit into it. 1/10+ 1/10 = 2/10 which is also 1/5 so it will take 2 days.

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If Tanya starts at one vertex of the pentagon and walks
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Step-by-step explanation:

That will be the perimeter.  Perimeter is the distance around.   Add all five sides, be sure to put the unit behind.

I need more information to help any further.

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3 years ago
a six sided number cube has faces with the numbers 1 through 6 marked on them. what is the probability that a number less than 3
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1/3 would be the answer
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Which value is a solution of the equation 2-8x=-6 ? ( 1point )
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<span>Here the given equation is : 2-8x= -6 By bring -6 on the Left Hand size and taking 8x on the Right Hand side, the equation become. 2+6= 8X 8= 8X X=8/8 X=1 Hence value of X is 1. Hence correct option is A.</span>
5 0
3 years ago
How can I solve 2x+y=-3
Vladimir79 [104]

Answer:

If solving for x then x = -3/2 +y/2

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4 0
3 years ago
A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0
3 years ago
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