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3 years ago

6

4y-1=15 written in x terms

1 answer:

Orlov [11]3 years ago

7 0

Answer:

y=7/2

Step-by-step explanation:

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Answer this question if cool

maxonik [38]

Answer:

The answer to your question is given below

Step-by-step explanation:

From the question given above, the mean score is 23.

Thus, we can obtain the distance from the mean (absolute deviation) by using the following formula:

Mean deviation = | mean – Score |

Mean = 23

For Score 21:

Mean deviation = | mean – Score |

Mean deviation = | 23 – 21 | = 2

For Score 22:

Mean deviation = | mean – Score |

Mean deviation = | 23 – 22 | = 1

For Score 28:

Mean deviation = | mean – Score |

Mean deviation = | 23 – 28 | = 5

For Score 29:

Mean deviation = | mean – Score |

Mean deviation = | 23 – 29 | = 6

SUMMARY:

Score >> Mean >> Absolute deviation

21 >>>>> 23 >>>>> 2

21 >>>>> 23 >>>>> 2

21 >>>>> 23 >>>>> 2

22 >>>>> 23 >>>>> 1

22 >>>>> 23 >>>>> 1

22 >>>>> 23 >>>>> 1

22 >>>>> 23 >>>>> 1

22 >>>>> 23 >>>>> 1

28 >>>>> 23 >>>>> 5

29 >>>>> 23 >>>>> 6

7 0

3 years ago

A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity afte

ELEN [110]

Answer:

410.32

Step-by-step explanation:

Given that the initial quantity, Q= 6200

Decay rate, r = 5.5% per month

So, the value of quantity after 1 month, $q_1 = Q- r \times Q$

$q_1 = Q(1-r)\cdots(i)$

The value of quantity after 2 months, $q_2 = q_1- r \times q_1$

$q_2 = q_1(1-r)$

From equation (i)

$q_2=Q(1-r)(1-r) \\\\q_2=Q(1-r)^2\cdots(ii)$

The value of quantity after 3 months, $q_3 = q_2- r \times q_2$

$q_3 = q_2(1-r)$

From equation (ii)

$q_3=Q(1-r)^2(1-r)$

$q_3=Q(1-r)^3$

Similarly, the value of quantity after n months,

$q_n= Q(1- r)^n$

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

$q_{48}=Q(1-r)^{48}$

Putting Q=6200 and r=5.5%=0.055, we have

$q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32$

Hence, the value of quantity after 4 years is 410.32.

4 0

3 years ago

Read 2 more answers

A rectangle has length 127.3 cm and width 86.5 cm, both correct to 1 decimal place. Calculate the upperbound and the lowerbound

Mnenie [13.5K]

Answer:

Correct to 1dp

127.3 cm = 127.0 cm

86.5 cm = 87.0 cm

Upper limits:

127.0 cm = 127.05 cm

87.0 cm = 87.05 cm

Lower Limits:

127.0 cm = 126.95 cm

87.0 cm = 86.95 cm

upper limit of perimeter of rectangle:

P = 2(l+w)

= 2(127.05 + 87.05)

= 2(214.1)

= 428.2 cm

lower limit of perimeter of rectangle:

P = 2(l+w)

= 2(126.95 + 86.95)

= 2(213.9)

= 427.8 cm

therefore;

$427.8 cm \leqslant perimeter < 428.2cm$

8 0

3 years ago

Three numbers that multiply to get to 24

mojhsa [17]

Answer: 1 x 24, 2 x 12, 3 x 8, and 4 x 6.

Step-by-step explanation: The factor pairs of 24 are: 1 x 24, 2 x 12, 3 x 8, and 4 x 6.

Hope This Helps

3 0

3 years ago

Read 2 more answers

Help please this is also do very very soon thank you!!!

stepan [7]

Answer:

700

Step-by-step explanation:

100:2

???:14

14*100

=1400

1400/2

=700

3 0

3 years ago

Read 2 more answers