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jolli1 [7]
3 years ago
15

Please help me with this question.​

Mathematics
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

Step-by-step explanation:

Hello

\sqrt{3^2+4*3+4}=\sqrt{9+12+4}=\sqrt{25}=5

So the limit is 5

thanks

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A chef got 7 bags of onions. The red onions came in bags of 8 and the yellow onions came in bags of 3. If the chef got a total o
frutty [35]

Answer:

\#\text{ of 8-onion bags: }4,\\\#\text{ of 3-onion bags: }3

Step-by-step explanation:

Let a be the number of bags with 8 onions and let b be the number of bags with 3 onions. We have the following system of equations:

\begin{cases}a+b=7,\\8a+3b=41\end{cases}

Subtracting b from both sides of the first equation, we get a=7-b. Substitute this into the second equation:

8(7-b)+3b=41,\\56-8b+3b=41,\\56-5b=41,\\-5b=-15,\\b=\boxed{3}

Therefore, the number of 8-onion bags is:

a=7-b,\\a=7-3,\\a=\boxed{4}

Thus, the chef got 4 8-onion bags and 3 3-onion bags.

5 0
3 years ago
From 12,000 graphing calculators produced by a​ manufacturer, an inspector selects a random sample of 550 calculators and finds
maria [59]

Answer:

131

Step-by-step explanation:

Using proportion,

If Out of 550 calculators = 6 are defective,

Then out of 12000 calculators = ? are defective

\frac{12000 calculators}{550calculators} * 6

= 21.8 * 6

= 130.8 ≈ 131

∴ Approximately 131 calculators are defective

3 0
3 years ago
If F^-1(x) if f(x) = (x-2)^2 for >2
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I don’t see the choices
6 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
What is the distance between the pair of points (7.2) and (10,6).
Viefleur [7K]

Answer:

(-13,-7)

Step-by-step explanation:

7 0
3 years ago
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