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Nezavi [6.7K]
3 years ago
9

A. 4+2(5)+(-3) Can you please give me the answer on this problem?

Mathematics
2 answers:
navik [9.2K]3 years ago
8 0
11 ishsbbehdhxhdhxhhxhx
a_sh-v [17]3 years ago
6 0
The answer is 11 :)
I know this is the correct answer because of PEMDAS.
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Y=99.99+23.75(x-1) as a Y=ax+b equation
Hatshy [7]

Answer:

y = 23.75x  + 76.24

Step-by-step explanation:

We want to rewrite

y = 99.99 + 23.75(x - 1)

in the form

y = ax + b

We expand to get:

y = 99.99 + 23.75x - 23.75

Regroup similar terms:

y = 23.75x - 23.75 + 99.99

Simplify:

y = 23.75x  + 76.24

This is now of the form

y = ax + b

where a=23.75 and b=76.24

6 0
3 years ago
Equalivent fraction of 2/3
LiRa [457]
2/3 could equal 4/6 , 6/9, 8/12 , 10/15 , 12/18 ETC
6 0
3 years ago
Read 2 more answers
The temperature outside is currently 10°F. It will fall at a constant rate of 3°F per hour. Which temperature will it be in 4 ho
Triss [41]

Answer:

T4 = -2°F

Step-by-step explanation:

Given the following data;

Current temperature = 10°F

Constant, K = 3°F

Time, t = 4 hours

In four (4) hours, we have;

T = kt

Substituting into the equation, we have;

T = 3 * 4

T = 12°F

T4 = current temp - new temp

T4 = 10°F - 12°F

T4 = -2°F

Therefore, the temperature it will be in 4 hours is - 2°F.

6 0
3 years ago
Which set of angle measures could form a triangle?
777dan777 [17]
Answer: C

Triangle=36+42+102=180
4 0
3 years ago
A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou
Vaselesa [24]

Answer:

The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let z be the

\displaystyle z = \frac{x - \mu}{\sigma}.

In this question, x = 5 and \sigma = 2. The equation becomes

\displaystyle z = \frac{5 - \mu}{2}.

To solve for \mu, the mean of this distribution, the only thing that needs to be found is the value of z. Since

The problem stated that P(X \le 5) = 69.85\% = 0.6985. Hence, P(Z \le z) = 0.6985.

The problem is that the normal distribution tables list only the value of P(0 \le Z \le z) for z \ge 0. To estimate  z from P(Z \le z) = 0.6985, it would be necessary to find the appropriate

Since P(Z \le z) = 0.6985 and is greater than P(Z \le 0) = 0.50, z > 0. As a result, P(Z \le z) can be written as the sum of P(Z < 0) and P(0 \le Z \le z). Besides, P(Z < 0) = P(Z \le 0) = 0.50. As a result:

\begin{aligned}&P(Z \le z)\\ &= P(Z < 0) + P(0 \le Z \le z) \\ &= 0.50 + P(0 \le Z \le z)\end{aligned}.

Therefore:

\begin{aligned}&P(0 \le Z \le z) \\ &= P(Z \le z) - 0.50 \\&= 0.6985 - 0.50 \\&=0.1985 \end{aligned}.

Lookup 0.1985 on a normal distribution table. The corresponding z-score is 0.52. (In other words, P(0 \le Z \le 0.52) = 0.1985.)

Given that

  • z = 0.52,
  • x =5, and
  • \sigma = 2,

Solve the equation \displaystyle z = \frac{x - \mu}{\sigma} for the mean, \mu:

\displaystyle 0.52 = \frac{5 - \mu}{2}.

\mu = 5 - 2 \times 0.52 = 3.96.

3 0
3 years ago
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