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3 years ago

Calculate the length of the diagonal AB

Mathematics

1 answer:

Vesna [10]3 years ago

6 0

Answer:

Please add a attachment.

Step-by-step explanation:

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I’m not sure where to begin

Dmitry [639]

(a)

we are given

the price of apple suddenly up by $0.75

Let's assume original price be 'p'

we are given

Sam bought 3 pounds of apple at the new price total is $5.88

so, new price = old price +0.75

price of 3 apples $=3(p+0.75)$

we are given that price is 5.88

so, we get equation as

$5.88=3(p+0.75)$ .......Answer

(b)

Since, original price is p

so, we have to solve for p

$5.88=3p+2.25$

$5.88-2.25=3p+2.25-2.25$

$3.63=3p$

$p=1.21$

so, original price is $1.21 per pound........Answer

8 0

3 years ago

Is this the right answer correct me if im wrong what is m

REY [17]

I think that is correct !!

5 0

2 years ago

For f(x)=4x+1 and g(x)=x^2-5, find (g/f)(x)

ddd [48]

Answer:

$(g/f)(x) = \frac{g(x)}{f(x)} = \frac{ {x}^{2} - 5 }{4x + 1}$

I hope I helped you^_^

3 0

3 years ago

Pls help 30 points 7th grade math

Vlada [557]

Z>7 I think that’s the answer let me know if it’s right or not

5 0

3 years ago

The variance can never bea.zerob.larger than the standard deviationc.negatived.smaller than the standard deviation

MrRa [10]

Answer:

c.negative

True, by definition since the variance take in count the differences around the mean squared, then we can't have a negative value for a sum of square values.

Step-by-step explanation:

We need to remember that the variance is a measure of dispersion for a dataset respect to a measure of central tendency called the mean. The mean is defined as:

$\bar x = \frac{\sum_{i=1}^n X_i}{n}$

And the population mean is given by:

$\mu= \frac{\sum_{i=1}^n X_i}{N}$

The population variance is defined by:

$\sigma^2 = \frac{\sum_{i=1}^n (X_i -\mu)^2}{N}$

And the sample variance is defined as:

$s^2= \frac{\sum_{i=1}^n (X_i -\bar x)}{n-1}$

Now if we analyze the options we have this:

a.zero

False, if all the values are the same then the mean would be the same to the values and the difference between each value and the mean would 0 and indeed the variance would be 0.

b.larger than the standard deviation

False, if the population standard deviation is $\sigma=2$ then the variance would be $\sigma^2 = 4$ and as we can see is higher than the standard deviation

c.negative

True, by definition since the variance take in count the differences around the mean squared, then we can't have a negative value for a sum of square values.

d.smaller than the standard deviation

False, if the population standard deviation is $\sigma=0.1$ then the variance would be $\sigma^2 = 0.01$ and as we can see is lower than the standard deviation

7 0

3 years ago