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1 year ago

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Tim wants to purchase a watch for $98.99. He has a coupon that will discount the watch by 10%. If sales tax is 7.5%, what will T

im pay at the register? (Round to the nearest penny)

1 answer:

PtichkaEL [24]1 year ago

7 0

Answer:

$95.77

Step-by-step explanation:

We are told that Tim wants to purchase a watch for $98.99.

He has a coupon that will discount the watch by 10%. So price of watch after using coupon will be,

$\text{Price of watch after discount}=98.99-(98.99*\frac{10}{100})$

$\text{Price of watch after discount}=98.99-(98.99*0.10)$

$\text{Price of watch after discount}=98.99-9.899=89.091$

Now let us find the price of watch after sales tax of 7.5%.

$\text{Price of watch after sales tax}=89.091+(89.091*\frac{7.5}{100})$

$\text{Price of watch after sales tax}=89.091+(89.091*0.075)$

$\text{Price of watch after sales tax}=89.091+6.681825$

$\text{Price of watch after sales tax}=95.772825\approx 95.77$

Therefore, Tim will pay $95.77 at the register.

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6 points Emily’s family needs to rent a moving truck to move their belongings to a different house. The rental cost for Trucks-A

Alexxx [7]

Answer/Step-by-step explanation:

Equation to represent the daily rental cost for each type of truck can be written as follows:

Daily rental cost for Trucks-A-Lot = 42 + 0.72m

Daily rental cost for Move-in-Truckers = 70 + 0.12m

Where, m = Emily's mileage

To determine the number of miles for which the truck cost the same amount, set both equations equal to each other and solve for m.

$42 + 0.72m = 70 + 0.12m$

Collect like terms

$0.72m - 0.12m = 70 - 42$

$0.6m = 28$

Divide both sides by 0.6

$\frac{0.6m}{0.6} = \frac{28}{0.6}$

$m = 46.7$

At approximately 47 miles, both trucks would cost the same amount.

Check:

Daily rental cost for Trucks-A-Lot = 42 + 0.72m

Plug in the value of x = 47

= 42 + 0.72(47) = $75.84 ≈ $76

Daily rental cost for Move-in-Truckers = 70 + 0.12m

Plug in the value of x = 47

= 70 + 0.12(47) = $75.64 ≈ $76

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2 years ago

Transversal CD←→ cuts parallel lines PQ←→ and RS←→ at points X and Y as shown in the diagram. If m∠CXP = 106.02°, what is m∠SYD?

Softa [21]

<span>m∠SYD = </span>106.02°
..............................................................

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The table shows how an elevator 500 feet above the ground is descending at a steady rate.

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We are given that the elevator is descending, so the coefficient of t must be negative. The elevator's initial position is 500 feet above the ground, so this is a positive value of +500. Therefore the only choice that fits these is h(t) = -5t + 500.

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A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

tankabanditka [31]

The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
<span>"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

</span> $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$
<span>
P(all 6 girls chosen)=35/286=0.12

B.
"</span>The probability that a randomly chosen team has 3 girls and 7 boys.<span>"

with the same logic as in A, the number of groups were all 7 boys are in, is

</span> $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$
<span>
so the probability is 20/286=0.07

C.
"</span>The probability that a randomly chosen team has either 4 or 6 boys.<span>"

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is </span>0.12.
<span>
case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

</span> $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$
<span>
the probability is 105/286=0.367

since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"</span><span>The probability that a randomly chosen team has 5 girls and 5 boys.</span><span>"

selecting 5 boys and 5 girls can be done in

</span> $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

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2 years ago

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Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random sample of

nikitadnepr [17]

Answer:

a) Reject H₀

b) [0.31; 3.35]

Step-by-step explanation:

Hello!

a) The objective of this example is to compare if the population means of the production rate of the assembly lines A, B and C. To do so the data of the production of each line were recorded and an ANOVA was run using it.

The study variable is:

Y: Production rate of an assembly line.

Assuming that the study variable has a normal distribution for each population, the observations are independent and the population variances are equal, you can apply a parametric ANOVA with the hypothesis:

H₀ μ₁= μ₂= μ₃

H₁: At least one of the population means is different from the others

Where:

Population 1: line A

Population 2: line B

Population 3: line C

α: 0.01

This test is always one-tailed to the right. The statistic is the Snedecor's F, constructed as the MSTr divided by the MSEr if the value of the statistic is big, this means that there is a greater variance due to the treatments than to the error, this means that the population means are different. If the value of F is small, it means that the differences between populations are not significant ( may differ due to error and not treatment).

The critical region is:

$F_{k-1;n-k; 1-\alpha } = F_{2;15; 0.99} = 6.36$

If F ≥ 3.36, the decision is to reject the null hypothesis.

Looking at the given data:

$F= \frac{MSTr}{MSEr}$ = 11.32653

With this value the decision is to reject the null hypothesis.

Using the p-value method:

p-value: 0.001005

α: 0.01

The p-value is less than the significance level, the decision is to reject the null hypothesis.

At a level of 5%, there is significant evidence to say that at least one of the population means of the production ratio of the assembly lines A, B and C is different than the others.

b) In this item, you have to stop paying attention to the production ratio of the assembly line A to compare the population means of the production ratio of lines B and C.

(I'll use the same subscripts to be congruent with part a.)

The parameter to estimate is μ₂ - μ₃

The populations are the same as before, so you can still assume that the study variables have a normal distribution and their population variances are unknown but equal. The statistic to use under these conditions, since the sample sizes are 6 for both assembly lines, is a pooled-t for two independent variables with unknown but equal population variances.

t= (X[bar]₂ - X[bar]₃) - ( μ₂ - μ₃) ~t $_{n_2+n_3-2}$

Sa√(1/n₂+1/n₃)

The formula for the interval is:

(X[bar]₂ - X[bar]₃) ± $t_{n_2+n_3-2; 1 - \alpha /2}$ * $Sa\sqrt{*\frac{1}{n_2} + \frac{1}{n_3} }$

$Sa^{2} = \frac{(n_2-1)*S_2^2+ (n_3-1)*S_3^2}{n_2+n_3-2}$

$Sa^{2} = \frac{(5*0.67)+ (5*0.7)}{6+6-2}$

$Sa^{2} = 0.685$

Sa= 0.827 ≅ 0.83

$t_{n_2+n_3-2;1-\alpha /2}= t_{10;0.995} = 3.169$

X[bar]₂ = 43.33

X[bar]₃ = 41.5

(43.33-41.5) ± 3.169 * $*0.83\sqrt{*\frac{1}{6} + \frac{1}{6} }$

1.83 ± 3.169 * 0.479

[0.31; 3.35]

With a confidence level of 99% you'd expect that the difference of the population means of the production rate of the assemly lines B and C.

I hope it helps!

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1 year ago