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Effectus [21]
2 years ago
8

Can you help me?Answer the question please​

Mathematics
1 answer:
Harlamova29_29 [7]2 years ago
5 0

Hey there! :D

1. c - 7 = 32

To solve this problem, add the difference and the subtrahend of the given equation.

32 + 7 = \boxed{\sf 39}

Checking:

To check, simply substitute the conclusion on the unknown quantity.

39 (c) - 7 = 32

Thus, the equation was correct which gives us the answer that:

\large \boxed{ \sf {\: c = 39}}

\begin{gathered} \\ \end{gathered}

2. r + 4 = 39

To solve this problem, subtract the sum and the addend of the given equation.

39 - 4 = \boxed{\sf 35}

Checking:

To check, simply substitute the conclusion on the unknown quantity.

35 (r) + 4 = 39

Thus, the equation was correct which gives us the answer that:

\large\boxed{ \sf {r \: = \: 35}}

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Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n
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Complete Question

Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n=150

Answer:

Normal sampling distribution can not be used

Step-by-step explanation:

From the question we are told that

    The  null hypothesis is  H_o  :  p =  0.015

     The  alternative hypothesis is    H_a  :  p  <  0.015

     

The  sample size is  n= 150

Generally in order to use  normal sampling distribution  

     The value  np  \ge  5

So  

         np =  0.015 * 150

         np =  2.25

Given that  np < 5   normal sampling distribution  can not be used

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3a, 5b, -6y, 10 hope this helps
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What is the probability of drawing a 3 or a 4 or a heart from a deck of cards?
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\Omega=\{2\spadesuit;\ 3\spadesuit;...;A\spadesuit;2\heartsuit ;\ 3\heartsuit;...;A\heartsuit;\ 2\diamondsuit;\ 3\diamondsuit;...;A\diamondsuit;\ 2\clubsuit;\ 3\clubsuit;...;A\clubsuit\}\\\\|\Omega|=52\\\\A=\{3\spadesuit;\ 4\spadesuit;\ 3\diamondsuit;\ 4\diamondsuit;\ 3\clubsuit;\ 4\clubsuit;\ 2\heartsuit;\ 3\heartsuit;\ 4\heartsuit;...;A\heartsuit\}\\\\|A|=2+2+2+13=19\\\\P(A)=\dfrac{|A|}{|\Omega|}\to P(A)=\dfrac{19}{52}

Answer:\ \dfrac{19}{52}
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A car moving at an initial velocity of 20 meters/second accelerates at the rate of 1.5 meters/second2 for 4 seconds. The car’s f
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Segments
Ipatiy [6.2K]

There are several ways two triangles can be congruent.

  • \mathbf{AC = BD}<em> congruent by SAS</em>
  • \mathbf{\angle ABC \cong \angle BAD}<em> congruent by corresponding theorem</em>

In \mathbf{\triangle AOL} and \mathbf{\triangle BOK} (see attachment), we have the following observations

1.\ \mathbf{AO = DO} --- Because O is the midpoint of line segment AD

2.\ \mathbf{BO = CO} --- Because O is the midpoint of line segment BC

3.\ \mathbf{\angle AOB =\angle COD} ---- Because vertical angles are congruent

4.\ \mathbf{\angle AOC =\angle BOD} ---- Because vertical angles are congruent

Using the SAS (<em>side-angle-side</em>) postulate, we have:

\mathbf{AC = BD}

Using corresponding theorem,

\mathbf{\angle ABC \cong \angle BAD} ---- i.e. both triangles are congruent

The above congruence equation is true because:

  1. <em>2 sides of both triangles are congruent</em>
  2. <em>1 angle each of both triangles is equal</em>
  3. <em>Corresponding angles are equal</em>

See attachment

Read more about congruence triangles at:

brainly.com/question/20517835

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2 years ago
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