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3 years ago

How to you solve 4(3-6a) = 36

2 answers:

5 0

4(3-6a)=36
4(3-6a)/4=36/4 divide by 4 on both sides
3-6a-3=9-3 subtract 3 on both sides
-6a/-6=6/-6 divide by -6 on both sides
<span>a=-1 solution</span>

sasho [114]3 years ago

3 0

A=-1 So there is the answeerrrrrr

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3 years ago

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3 years ago

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 436.0 gram setting. It

Vilka [71]

Answer:

We conclude that the machine is under filling the bags.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 436.0 gram

Sample mean, $\bar{x}$ = 429.0 grams

Sample size, n = 40

Alpha, α = 0.05

Population standard deviation, σ = 23.0 grams

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 436.0\text{ grams}\\H_A: \mu < 436.0\text{ grams}$

We use one-tailed(left) z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{429 - 436}{\frac{23}{\sqrt{40}} } = -1.92$

Now, $z_{critical} \text{ at 0.05 level of significance } = -1.64$

Since,

$z_{stat} < z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis. Thus, we conclude that the machine is under filling the bags.

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3 years ago

Which is equivalent to (3 + 2i)(3 + 2i)

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4 years ago

The weight of strawberries follows a normal distribution with a mean weight of 12 grams and a standard deviation of 2.5 grams. I

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This is a problem of Standard Normal distribution.

We have mean= 12 grams
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First we convert 8.5 to z score. 8.5 converted to z score for given mean and standard deviation will be:

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So, from standard normal table we need to find the probability of z score to be less than -1.4. The probability comes out to be 0.0808

Thus, the <span>probability that the strawberry weighs less than 8.5 grams is 0.0808</span>

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3 years ago