Shane has nine notes in his wallet, three are $50 and the rest are $20 notes. Write the ratio of $50 notes to $20 dollar notes.

The average number of traffic tickets issued in a city on any given day

Anton [14]

Answer:

The most tickets were written on Saturday .On Saturday 325 tickets were issued

Step-by-step explanation:

The average number of traffic tickets issued in a city on any given day Sunday-Saturday can be approximated by

$T(x) = -6x^2 + 84x + 37$

Where x represents the number of days after Sunday

T(x) represents the number of traffic tickets issued.

Sunday = x=0

Monday = x=1

Tuesday = x=2

Wednesday = x=3

Thursday = x =4

Friday = x=5

Saturday = x=6

Substitute x= 0

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(0)^2 + 84(0) + 37\\T(x)=37$

On Sunday 37 tickets were issued

Substitute x= 1

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(1)^2 + 84(1) + 37\\T(x)=115$

On Monday 115 tickets were issued

Substitute x= 2

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(2)^2 + 84(2) + 37\\T(x)=181$

On Tuesday 181 tickets were issued

Substitute x= 3

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(3)^2 + 84(3) + 37\\T(x)=235$

On Wednesday 235 tickets were issued

Substitute x= 4

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(4)^2 + 84(4) + 37\\T(x)=277$

On Thursday 277 tickets were issued

Substitute x= 5

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(5)^2 + 84(5) + 37\\T(x)=307$

On Friday 307 tickets were issued

Substitute x= 6

$T(x) = -6x^2 + 84x + 37\\T(x) = -6(6)^2 + 84(6) + 37\\T(x)=325$

On Saturday 325 tickets were issued

Hence the most tickets were written on Saturday .On Saturday 325 tickets were issued

6 0

3 years ago

1. Kalena was asked to prove ifx(x - 1)(x + 1) = x3 - X represents a polynomial identity. She

Verdich [7]

Answer:

C. Kalena made a mistake in Step 3. The justification should state: -x²

+ x²

Step-by-step explanation:

Given the function x(x - 1)(x + 1) = x3 - X

To justify kelena proof

We will need to show if the two equations are equal.

Starting from the RHS with function x³-x

First we will factor out the common factor which is 'x' to have;

x(x²-1)

Factorising x²-1 using the difference of two square will give;

x(x+1)(x-1)

Note that for two real number a and b, the expansion of a²-b² using difference vof two square will give;

a²-b² = (a+b)(a-b) hence;

Factorising x²-1 using the difference of two square will give;

x(x+1)(x-1)

Factorising x(x+1) gives x²+x, therefore

x(x+1)(x-1) = (x²+x)(x-1)

(x²+x)(x-1) = x³-x²+x²-x

The function x³-x²+x²-x gotten shows that kelena made a mistake in step 3, the justification should be -x²+x² not -x-x²

6 0

3 years ago

(x+7)^2+(y+8)^2=64 center and radius

Pie

(-7,-8) is the center and 8 is the radius.

6 0

3 years ago

Read 2 more answers

The aquarium has 198 guppies, angelfish and catfish altogether.

BabaBlast [244]

Answer:

Number of catfish = 28

Step-by-step explanation:

Let the number of catfish = c

Number of angle fish = c + 6

Number of guppies = 4*(c + 6) = 4c + 4*6 = 4c + 24

Total fishes = 198

c + c + 6 + 4c + 24 = 198 {Combine like terms}

c + c +4c + 6 + 24 = 198

6c + 30 = 198 {Subtract 30 from both sides}

6c = 198 - 30

6c = 168

c = 168/6 {Divide both sides by 6}

c = 28

5 0

3 years ago

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0

3 years ago