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3 years ago

Combine like terms: 7x^2−5x^3+8x^2−6x^3

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

7 0

Final answer is:

15x^2-11x^3

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Solve when m = 7 and k = 2 6m- 3k A.30 B.21 C.36 D.35

NARA [144]

Agree with the one at the top it’s 36

6 0

2 years ago

Read 2 more answers

Based on 2017 sales, the six top-selling compact cars are the Honda Civic, Toyota Corolla, Nissan Sentra, Hyundai Elantra, Chevr

fgiga [73]

Answer:

Pvalue = 0.020335

X² = 14.99

CONCLUSION: THERE IS DIFFERENCE IN MARKET SHARES

Step-by-step explanation:

H0 : there is no difference in market shares

H1: there is difference in market shares

Chisquare (X²) = (observed - Expected)²/ expected

Honda Civic - 0.20 * 400 = 80 : (98-80)^2 /80 = 4.05

Toyota Corolla - 0.17 * 400 = 68 : (72-68)²/68 = 0.235

Nissan Sentra - 0.12 * 400 = 48 : (54-48)²/48 = 0.75

Hyundai Elantra - 0.10 *400 = 40 : (44-40)²/40 = 0.4

Chevrolet Cruze - 0.10 * 400 = 40 : (42-40)²/40 = 0.1

Ford Focus - 0.08 * 400 = 32 : (25-32)²/32 = 1.531

Others - 0.23 * 400 = 92 : (65-92)²/92 = 7.924

X² = (4.05+0.235+0.75+0.4+0.1+1.531+7.924) = 14.99

The obtain the Pvalue ;

We use the Pvalue from Chisquare calculator :

Pvalue(14.99, 0.05, df= 6) = 0.020335

Decision region :

Reject H0 ; if Pvalue < α

THERE IS DIFFERENCE IN MARKET SHARES

Market share differences:

* 98/400 = 0.245,

0.245-0.20 = 0.045

* 72/400 = 0.18,

0.18 - 0.17 = 0.01

* 54/400 = 0.135,

0.135 - 0.12 = 0.015

* 44/400 = 0.11,

0.11 - 0.10 = 0.01

* 42/400 = 0.105

0.105 - 0.10 = 0.005

* 25/400 = 0.0625

0.0625-0.08 = -0.0175

* 65/400 = 0.1625

0.1625 - 0.23 = -0.0675

5 0

2 years ago

Is 80 homework and 74 exit tickets equivalent?

Elden [556K]

No. If one student each handed in their homework but six did not hand in an exit ticket.

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2 years ago

S What is the value of PS in the triangle above? (A) 5/2 (B) 10 (C) 11 (D) 13

makkiz [27]

Answer:

Step-by-step explanation:

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2 years ago

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$