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3 years ago

(3a3 − 5b3) + _ a3 + _b3 = (a3 + b3).

1 answer:

7 0

3a^3-5b^3+xa^3+yb^3=1a^3+1b^3
replace a^3 with z
replace b^3 with v (to make it easier to see)
3z-5v+xz+yv=z+v
3z+xz+yv-5v=1z+v

the terms match them
3z+xz=1z
divide both sides by z
3+x=1
minus 3
x=-2

yv-5v=1v
dividie both sides by v
y-5=1
addd 5 to both sides
y=6

(3a^3-5b^3)+ -2 a^3 +6 b^3=(a^3+b^3)

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When I sit and watch my students take exams, I often think to myself "I wonder if students with bright calculators are impacted

hodyreva [135]

Answer:

There is a difference between the two means.

Step-by-step explanation:

The hypothesis can be defined as:

H₀: The mean exam scores of my SAT 215 students with colorful calculators are same as the mean scores of my STA 215 students with plain black calculators, i.e. μ₁ - μ₂ = 0.

Hₐ: The mean exam scores of my SAT 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators, i.e. μ₁ - μ₂ ≠ 0.

Assume that the significance level of the test is, α = 0.05. Also assuming that the population variances are equal.

The decision rule:

A 95% confidence interval for mean difference can be used to determine the result of the hypothesis test. If the 95% confidence interval contains the null hypothesis value, i.e. 0 then the null hypothesis will not be rejected.

The 95% confidence interval for mean difference is:

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

Compute the pooled standard deviation as follows:

$S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}} {n_{1}+n_{2}-2}}}=\sqrt{\frac{(49-1)(4.7)^{2}+(38-1)(5.7)^{2}}{49+38-2}}=5.16$

The critical value of t is:

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (49+38-2)}=t_{0.025, 85}=1.984$

*Use a t-table.

Compute the 95% confidence interval for mean difference as follows:

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

$=(84-87)\pm 1.984\times 5.16\times \sqrt{\frac{1}{49}+\frac{1}{38}}$

$=-3\pm 2.133\\=(-5.133, -0.867)\\\approx(-5.13, -0.87)$

The 95% confidence interval for mean difference is (-5.13, -0.87).

The confidence interval does not contains the value 0. This implies that the null hypothesis will be rejected at 5% level of significance.

Hence, concluding that the mean exam scores of my STA 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators.

7 0

4 years ago

102,000 is 8% of what number? HELP!!!!!!!!!!

dusya [7]

Answer:

1275000

Step-by-step explanation:

102000/0.08=1275000

5 0

3 years ago

Read 2 more answers

The sum of the quantity x-3 and y

exis [7]

For this case we have the following expressions:
x-3
Y
Adding both expressions we have:
(x-3) + (y)
Rewriting we have:
x - 3 + y
y + x - 3
Answer:
The sum of the quantity x-3 and y is:
y + x - 3

7 0

4 years ago

Harry's mother makes cakes for a local resturant. She buys flour and sugar in large amounts. She has 157.86 pounds of flour and

dolphi86 [110]

Should last up to 10 days for both

3 0

3 years ago

Read 2 more answers

Measure of a verticle angle is 7x+182 and another angle is 9x+194 solve for x

disa [49]

The measures of 2 vertical angles are always equal, so we have:

7x+182=9x+192.

Rearranging, we have 182-192=9x-7x, which simplifies to -12=2x.

Thus, x=-6.

This means that the angles are:

i) 7x+182=7(-6)+182=182-42= 140(degrees).

ii) 9x+194=9(-6)+194=194-54=140 (degrees, as expected.)

Answer: x=-6

4 0

3 years ago

(3a3 − 5b3) + _____ a3 + _____b3 = (a3 + b3).

(3a3 − 5b3) + _ a3 + _b3 = (a3 + b3).