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3 years ago

The price of broccoli is $1.25 per pound. Let Y be the total cost in dollars of buying x pounds of broccoli

Mathematics

2 answers:

Minchanka [31]3 years ago

6 0

4. Y= 1.25x
Total cost= 1.25 x pounds of broccoli

Katarina [22]3 years ago

3 0

Answer:

4. y=1.25x

Step-by-step explanation:

y is the total cost

x are the pounds which will be multiplied to the price of the broccoli

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Hi please help me.. will give brainlest

chubhunter [2.5K]

Answer:

Volume = l*w*h

Equation: 1620 = 9*9*h

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h=20 cm

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2 years ago

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(-7)-(-44)/22+((-4)*59) with calculation

Pavel [41]

Hi,

$\frac{ ( - 7) - ( - 44)}{22 + (( - 4) \times 59)}$

$\frac{ - 7 - ( - 44)}{22 + (( - 4) \times 59}$

$\frac{ - 7 + 44}{22 + (( - 4) \times 59}$

$\frac{ - 7 + 44}{22 + ( - 236)}$

$\frac{37}{22 + ( - 236)}$

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3 years ago

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The realtor and her clients do not know the average home sale price for all of Guelph (500 was actually just a guess). However,

strojnjashka [21]

Answer:

a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=400$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=80$ represent the population standard deviation

n=15 represent the sample size

2) Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

Now we have everything in order to replace into formula (1):

$400-2.58\frac{80}{\sqrt{15}}=346.708$

$400+2.58\frac{80}{\sqrt{15}}=453.292$

So on this case the 99% confidence interval would be given by (346.708;453.292)

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

We need to find the degrees of freedom first df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that $t_{\alpha/2}=2.98$