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otez555 [7]
3 years ago
7

Please simplify this expression. 1 + 4.25n + 32p – 3 + (–2p) + 54n

Mathematics
2 answers:
Colt1911 [192]3 years ago
6 0
58.25n+30p-2 hope this helps
nata0808 [166]3 years ago
5 0

Answer:

58.25n+30p-2

Step-by-step explanation:

Given the expression as;

1 + 4.25n + 32p – 3 + (–2p) + 54n

collect like terms

1-3+4.25n+54n+32p-2p

-2+58.25n+30p

58.25n+30p-2

I have copied one brainly user's answer, sorry for that...

However, hope it helps you!!!

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Use the t-distribution to find a confidence interval for a difference in means μ1-μ2 given the relevant sample results. Give the
Serggg [28]

Answer:

(a) The best estimate of \mu_{1}-\mu_{2} is 13.2.

(b) The margin of error is 5.30.

(c) The 95% confidence interval for the difference between two means is (7.90, 18.50).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the difference between population means using a <em>t</em>-interval is:

CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)} \sqrt{S_{p}^{2}[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}

(a)

A point estimate of a parameter (population) is a distinct value used for the estimation the parameter (population). For instance, the sample mean \bar x is a point-estimate of the population mean μ.

Similarly the point estimate of the difference between two means is:

\bar x_{1}-\bar x_{2}

Compute the point estimate of \mu_{1}-\mu_{2} as follows:

E(\mu_{1}-\mu_{2})=\bar x_{1}-\bar x_{2}\\=79.0-65.8\\=13.2

Thus, the best estimate of \mu_{1}-\mu_{2} is 13.2.

(b)

Compute the pooled variance as follows:

S_{p}^{2}=\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}=\frac{(35-1)10.5^{2}+(20-1)7.2^{2}}{35+20-2}=89.311

Compute the critical value of <em>t</em> as follows:

t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (35+20-2)}=t_{0.025, 53}=2.00

*Use a <em>t</em>-table.

Compute the margin of error as follows:

MOE=t_{\alpha/2, (n_{1}+n_{2}-2)} \sqrt{S_{p}^{2}[\frac{1}{n_{1}}+\frac{1}{n_{2}}]}}

          =2.00\times\sqrt{89.311\times [\frac{1}{35}+\frac{1}{20}]} \\=5.298\\\approx5.30

Thus, the margin of error is 5.30.

(c)

Compute the 95% confidence interval for the difference between two means as follows:

CI=(\bar x_{1}-\bar x_{2})\pm MOE

      =13.2\pm 5.298\\=(7.902, 18.498)\\\approx (7.90, 18.50)

Thus, the 95% confidence interval for the difference between two means is (7.90, 18.50).

4 0
3 years ago
A solution of salt and water contains 70 grams of water per 140 milliliters of the solution. If 1 mole of water weighs 16 grams,
Mama L [17]

Answer:

1.09 moles.

Step-by-step explanation:

There is 70g of water in 140ml of solution.

This means there are 17.5g of water in 35ml of solution. (Divide by 4)

In 17.5g of water, there are 17.5/16 = 1.09 moles of water.

7 0
3 years ago
Write the product in its simplest form:<br> 8y^7*6y^7
valkas [14]
\bf 8y^7\cdot 6y^7\implies 8\cdot 6y^7y^7\implies 48y^{7+7}\implies 48y^{14}
6 0
3 years ago
Referring to the figure, find the value of x
lilavasa [31]

Answer:

x = 23

Step-by-step explanation:

3x + 3 + x + 85 = 180

4x + 88 = 180

4x = 92

x = 23

Hopefully this helps!

Brainliest please?

8 0
3 years ago
One coin is flipped six times. What is the probability of flipping two heads and four tails?
Scilla [17]

Answer:

1/6 x 1/6 = 1/36

1/36 is our probability of getting two heads and four tails.

4 0
3 years ago
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