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AleksandrR [38]
2 years ago
11

On a certain map, 2.5 inches represents 15 miles Bay City and green glitter 4 inches apart on the map what is the actual distanc

e between Bay City and Greenwood
Mathematics
1 answer:
Helga [31]2 years ago
5 0
The actual distance is 24 miles

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u is at (1,-2)

V is at (-6,-6)

 using distance formula it is about 8.06 long

 so Answer is C

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2 years ago
To improve his basketball skills, Troy decides to practice six times as many free throws and four times as many jump shots every
noname [10]
The answer is B. 6f+4j
6 0
3 years ago
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Solve for X <br><br>Explain if possible :)
sweet-ann [11.9K]

The two labeled angles are alternate interior angles, and as such, they are the same.

From this result you can build the equation

15x-2 = 13x+2

and solve it for x: subtract 13x from both sides to get

2x-2=2

and add 2 to both sides to get

2x = 4 \implies x = 2

Check: if we plug the value we found we have

13\cdot 2 + 2 = 26+2 = 28 = 15\cdot 2 - 2 = 30-2

So the angles are actually the same, as requested.

4 0
3 years ago
Ashley bought 4 packages of juice boxs in each package. She. gave 2 juice boxs to each of her friends. How many juice boxes does
Harlamova29_29 [7]
I am assuming there are more than one juice box in a package so if the amount of juice boxes in each package is X and there are four packages 4X.
So 4X - 2= Answer

If we make it so that there are 6 juice boxes in each package and use the same method this is what we'll get:
24 - 2= 22

22 would be the juice boxes left over.

I hope this answers your question, if it doesn't use the same method with the number of juice boxes in a package.
8 0
3 years ago
The nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium. To determine if the
lora16 [44]

Answer:

We conclude that the sodium content is same as what the nutrition label states.

Step-by-step explanation:

We are given that the nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium.

The FDA randomly selects 40 packages of Oriental Spice Sauce and determines the sodium content. The sample has an average of 1088.64 milligrams of sodium per package with a sample standard deviation of 234.12 milligrams.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average sodium content.</em></u>

So, Null Hypothesis, H_0 : \mu = 1100 milligrams      {means that the sodium content is same as what the nutrition label states}

Alternate Hypothesis, H_A : \mu \neq 1100 milligrams      {means that the sodium content is different from what the nutrition label states}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about population standard deviation;

                    T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}  ~ t_n_-_1

where, \bar X = sample average sodium content = 1088.64 milligrams

            s = sample standard deviation = 234.12 milligrams

            n = sample of packages of Oriental Spice Sauce = 40

So, <u><em>test statistics</em></u>  =  \frac{1088.64-1100}{\frac{234.12}{\sqrt{40}}}  ~ t_3_9

                              =  -0.307

The value of z test statistics is -0.307.

<em>Since, in the question we are not given the level of significance so we assume it to be 5%. </em><em>Now, at 0.05 significance level the t table gives critical values of -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test.</em><em> </em>

<em>Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which </em><em><u>we fail to reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the sodium content is same as what the nutrition label states.

3 0
3 years ago
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