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Vanyuwa [196]
3 years ago
6

The graph of y=x^2 is reflected in the x-axis and translated 3 units right and 2 units up. Write an equation for the function in

vertex form and in standard form.
Mathematics
1 answer:
Anna [14]3 years ago
8 0
Y= (x-3)^2 +2 I hope I helped :)
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Find the value of x.<br><br><br> A. 4.3<br> B. 2.7<br> C. 11.6<br> D. 13.9
kakasveta [241]

Answer:

<h2><u>[D] 13.9</u></h2>

Explanation:

  • <em>Pythagorean theorem: a² + b² = c²</em>
  • <em>Solve for hypotenuse (side x) using: c = √a² + b²</em>

12.8² + 5.3² = 191.93

√191.93

= 13.8538803229

<em>Round the answer</em>

13.9

6 0
3 years ago
HELP I NEED TO FIND THIS ANSWER PLEASE !!!
LUCKY_DIMON [66]
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7 0
3 years ago
A researcher plants 22 seedlings. After one month, independent of the other seedlings, each seedling has a probability of 0.08 o
Andrews [41]

Answer:

E(X₁)= 1.76

E(X₂)= 4.18

E(X₃)= 9.24

E(X₄)= 6.82

a. P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= 0.00022

b. P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= 0.000001

c. P(X₁≤2) = 0.7442

Step-by-step explanation:

Hello!

So that you can easily resolve this problem first determine your experiment and it's variables. In this case, you have 22 seedlings (n) planted and observe what happens with the after one month, each seedling independent of the others and has each leads to success for exactly one of four categories with a fixed success probability per category. This is a multinomial experiment so I'll separate them in 4 different variables with the corresponding probability of success for each one of them:

X₁: "The seedling is dead" p₁: 0.08

X₂: "The seedling exhibits slow growth" p₂: 0.19

X₃: "The seedling exhibits medium growth" p₃: 0.42

X₄: "The seedling exhibits strong growth" p₄:0.31

To calculate the expected number for each category (k) you need to use the formula:

E(XE(X_{k}) = n_{k} * p_{k}

So

E(X₁)= n*p₁ = 22*0.08 = 1.76

E(X₂)= n*p₂ = 22*0.19 = 4.18

E(X₃)= n*p₃ = 22*0.42 = 9.24

E(X₄)= n*p₄ = 22*0.31 = 6.82

Next, to calculate each probability you just use the corresponding probability of success of each category:

Formula: P(X₁, X₂,..., Xk) = \frac{n!}{X_{1}!X_{2}!...X_{k}!} * p_{1}^{X_{1}} * p_{2}^{X_{2}} *.....*p_{k}^{X_{k}}

a.

P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= \frac{22!}{3!4!6!} * 0.08^{3} * 0.19^{4} * 0.42^{6}\\ = 0.00022

b.

P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= \frac{22!}{5!5!7!} * 0.08^{5} * 0.19^{5} * 0.31^{7}\\ = 0.000001

c.

P(X₁≤2) = \frac{22!}{0!} * 0.08^{0} * (0.92)^{22} + \frac{22!}{1!} * 0.08^{1} * (0.92)^{21} + \frac{22!}{2!} * 0.08^{2} * (0.92)^{20} = 0.7442

I hope you have a SUPER day!

8 0
3 years ago
Grade 4c left their school at7:30 am for their trip to the market. They return at 10:45 am how much time did they spend on the t
SOVA2 [1]
3:15

Step by step explanation:

10:45
-7:30
———
3:15
6 0
2 years ago
The temperature in degrees Fahrenheit was recorded every two hours starting at midnight on the first day of summer. The data is
serg [7]
I’m pretty sure it’s line graph, if not that, bar graph
Hope this helps :D
6 0
3 years ago
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