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3 years ago

Help me please i need your help

Mathematics

2 answers:

Lunna [17]3 years ago

7 0

Answer:

$(3x + 4)^{2}$

Step-by-step explanation:

I think of a number = x

multiply by 3 = 3x (basically 3 x X)

add 4 = 3x + 4

square the result got from X x 3 + 4 = $(3x + 4)^{2}$

marishachu [46]3 years ago

4 0

Answer:

$\large\boxed{\boxed{\underline{\underline{\maltese{\pink{\pmb{\sf{\: SOLUTION: \: \: \: \: (3x + 4)^{2}}}}}}}}}$

Step-by-step explanation:

The given statement is;: 'I think of a number, multiply it by 3, add 4 & square the result.'

At first, let's take the unknown number as 'x' . (asked in the question).

So, the number we think about = x

Now, we need to multiply it by 3. So,

3 * x = 3x

In the next part, we need to add 4 to it so,

3x + 4

Now, by squaring the result, we'll get the expression as,

<u>(3x + 4)²</u>

Hope it helps!

$\mathfrak{Lucazz}$

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the sum of 4 consecutive even numbers is 92 . Write an equation in which x represents the smallest of these 4 numbers . Then sol

Svet_ta [14]

Let x = the first even number
x + 2 the second
x + 4 the third
x + 6 the fourth
Add together.
x + x + 2 + x + 4 + x + 6 = 92
Combine like terms.
4x + 12 = 92
Subtract 12 from both sides
4x = 80
Divide both sides by 4
x = 20

8 0

4 years ago

Suppose 244 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea. Use a 0.01 significance

vodka [1.7K]

Answer:

$z=\frac{0.209 -0.2}{\sqrt{\frac{0.2(1-0.2)}{244}}}=0.351$

$p_v =P(z>0.351)=0.363$

So the p value obtained was a very high value and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of subjects with nauseas is not higher than 0.2 or 20% at 1% of significance.

Step-by-step explanation:

Data given and notation

n=244 represent the random sample taken

X=51 represent the subjects with nausea

$\hat p=\frac{51}{244}=0.209$ estimated proportion of subjects with nausea

$p_o=0.2$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.2.:

Null hypothesis: $p \leq 0.2$

Alternative hypothesis: $p > 0.2$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.209 -0.2}{\sqrt{\frac{0.2(1-0.2)}{244}}}=0.351$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>0.351)=0.363$

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