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3 years ago

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

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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

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3 years ago

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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.

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3 years ago

Which set of angle measures could form a triangle?

777dan777 [17]

Answer: C

Triangle=36+42+102=180

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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: