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lord [1]
2 years ago
15

The rate of planting the grass is ` x per square meter. Find the cost of planting the grass on a

Mathematics
1 answer:
kotegsom [21]2 years ago
3 0

Step1: Find the area of the triangular lawn

             Given, base of the triangle is y metres and the height is z metres

             Area of the triangle =

2

1

​

× base × height

             Therefore, area of the triangular lawn =

2

1

​

yz metre  

2

.

Step 2:Find the cost of planting the grass

             Rate of planting the grass is Rs. x per square metre.

             Therefore, the cost of planting the grass on a triangular lawn =cost per square meter × area of the triangular lawn

              =x×

2

1

​

yz=

2

1

​

xyz

Hence, the cost of planting the grass on a triangular lawn whose base is y metres and  height is z metres is Rs.  

2

1

​

x y z.

Step-by-step explanation:

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vodomira [7]

Answer:

-3.75

Step-by-step explanation:

the opposite changes the sign from positive to negative, so the opposite of 3.75 would be -3.75

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zlopas [31]

0.46

Step-by-step explanation:

23÷50= 0.46

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This is a GCSE maths question which i don’t understand.
Tpy6a [65]

Answer:

15

Step-by-step explanation:

7 0
3 years ago
QUESTION 1 The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards.Assume that the driving
GarryVolchara [31]

Answer:

P(X>300)

And we can find this probability with the complement rule:

P(X>300)= 1-P(X

Step-by-step explanation:

For this case we define the random variable X ="driving distance for the top 100 golfers on the PGA tour" and we know that:

X \sim Unif (a=284.7, b=310.6)

And for this case the probability density function is given by:

f(x) =\frac{1}{310.6 -284.7} =0.0386 , 284.7 \leq X \leq 310.6

And the cumulative distribution function is given by:

F(x) =\frac{x-284.7}{310.6-284.7} , 284.7 \leq X \leq 310.6

And we want to find this probability:

P(X>300)

And we can find this probability with the complement rule:

P(X>300)= 1-P(X

6 0
3 years ago
A manufacturer of detergent claims that the contents of boxes sold weigh on average at least 20 ounces. The distribution of weig
jenyasd209 [6]

Answer:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

Step-by-step explanation:

Information provided

\bar X=19.87 represent the sample mean

\sigma=0.4 represent the population deviation

n=25 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0.01 represent the significance level

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:  

Null hypothesis:\mu \geq 20  

Alternative hypothesis:\mu < 20  

The statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Replacing the info given we got:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

7 0
3 years ago
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