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3 years ago

17d+32a=176 what is d?

Mathematics

1 answer:

Vlad [161]3 years ago

4 0

Minus 32a from both sides
17d=176-32a
divide both sides by 17
d= $\frac{176-32a}{17}$

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Jeri finds a pile of money with at least $\$200$. If she puts $\$50$ of the pile in her left pocket, gives away $\frac23$ of the

Marizza181 [45]

Answer:

The possible values of the number of dollars in the original pile of money is ≥ $200 but < $350

Step-by-step explanation:

Here we have, pile of money ≥ $200

Amount in put the left pocket = $50

Fraction given away = 2/3 of rest of pile ≥ 2/3×150 ≥ $100

Amount put in right pocket = ≥ $150 - $100 ≥ $50

Total amount remaining with Jeri = $50 +≥ $50 ≥ $100

Also original pile - $200 < $100

Therefore where maximum amount given away to have more money = $200 we have

2/3× (original pile - 50) = $200

Maximum amount for original pile = $350

Therefore the possible values of the number of dollars in the original pile of money is ≥ $200 but < $350.

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3 years ago

Is it more efficient to solve -3x+y=1 for x?why or why notes

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Answer:

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Step-by-step explanation:

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3 years ago

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34kurt

Answer:

b / (b/100×b)

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Step-by-step explanation:

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2 years ago

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Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) $t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

e) $p_v =P(t_{118}$

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

$n_{O}=60$ sample size for Orange Coast

$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis: $\mu_{O} \geq \mu_{C}$

Alternative hypothesis: $\mu_{O} < \mu_{C}$

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

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Paha777 [63]

Answer:

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Step-by-step explanation:

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