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3 years ago

I NEED HELP ASAP PLSSS

Mathematics

2 answers:

Triss [41]3 years ago

5 0

Answer:

y = x^2 - 4x - 5

y = (x + 1)(x - 5)

Lena [83]3 years ago

4 0

Answer:

$Standard \: form :\:y= {x}^{2} - 4x - 5\\Factored\: from:\:y = (x + 1) (x - 5)$

Step-by-step explanation:

$y = (x - 2)^{2} - 9 \\ y= {x}^{2} - 4x + 4 - 9 \\ \red{ \boxed{ \bold{y= {x}^{2} - 4x - 5}}} \\ is \: in \: standard \: form \\ \\ y = (x - 2)^{2} - 9 \\ y = (x - 2)^{2} - {3}^{2} \\ y = \{(x - 2) + 3\} \{(x - 2) - 3 \} \\ y = \{x - 2 + 3\} \{x - 2- 3 \} \\ \purple{ \boxed{ \bold{y = (x + 1) (x - 5)}}} \\ is \: in \: factored \: form$

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3 years ago

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The nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium. To determine if the

lora16 [44]

Answer:

We conclude that the sodium content is same as what the nutrition label states.

Step-by-step explanation:

We are given that the nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium.

The FDA randomly selects 40 packages of Oriental Spice Sauce and determines the sodium content. The sample has an average of 1088.64 milligrams of sodium per package with a sample standard deviation of 234.12 milligrams.

Let $\mu$ = average sodium content.

So, Null Hypothesis, $H_0$ : $\mu$ = 1100 milligrams {means that the sodium content is same as what the nutrition label states}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 1100 milligrams {means that the sodium content is different from what the nutrition label states}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample average sodium content = 1088.64 milligrams

s = sample standard deviation = 234.12 milligrams

n = sample of packages of Oriental Spice Sauce = 40

So, test statistics = $\frac{1088.64-1100}{\frac{234.12}{\sqrt{40}}}$ ~ $t_3_9$

= -0.307

The value of z test statistics is -0.307.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 0.05 significance level the t table gives critical values of -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the sodium content is same as what the nutrition label states.

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3 years ago

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