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3 years ago

The amount of money Irene earns for walking dogs is shown in the table.

Mathematics

2 answers:

sergeinik [125]3 years ago

8 0

Answer:$5.25

Step-by-step explanation:

is easy just do 10.50 divided by 2 and your answer is $5.25

sdas [7]3 years ago

3 0

Answer:

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Step-by-step explanation:

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Need help ASAP pls thx

Mazyrski [523]

$\\ \sf\longmapsto RElative\:Frequency(Green)$

$\\ \sf\longmapsto RF(Green)=\dfrac{28}{14+28+28}$

$\\ \sf\longmapsto RF(Green)=\dfrac{28}{70}$

$\\ \sf\longmapsto RF(Green)=\dfrac{2}{5}$

5 0

3 years ago

Read 2 more answers

Fill in the blank. _______ are sample values that lie very far away from the majority of the other sample values. Centers Distr

cestrela7 [59]

Answer:

Outliers are sample values that lie very far away from the majority of the other sample values. Centers Distributions Outliers Frequencies are sample values that lie very far away from the majority of the other sample values.

Step-by-step explanation:

4 0

3 years ago

!!! PLEASE ANSWER NOW!!! <br>A square has side length 4 cm. What is its area?<br>

almond37 [142]

Answer:

16 cm

Step-by-step explanation:

4 × 4 = 16 cm

I hope this helps!

8 0

3 years ago

Read 2 more answers

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

1.) Simplify: y = -8 + 6(x-1)

swat32

8+6=14. (X-1=1x)

14•1x

6 0

3 years ago