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3 years ago

Answer questions a) and b)

Mathematics

1 answer:

Mariulka [41]3 years ago

4 0

Answer:

$x\hookrightarrow -1\hookrightarrow 0\hookrightarrow 1\hookrightarrow 2\hookrightarrow 3\hookrightarrow 4\hookrightarrow 5$

$y\hookrightarrow 5\hookrightarrow 0\hookrightarrow -3\hookrightarrow -4\hookrightarrow -3\hookrightarrow 0\hookrightarrow 5$

The curve (B) matches the graph of y=x²-4x.

$----------\\Hope \ it\;helps$

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If 5+20-22-3x = 10-2-2x + 5, what is the value of x? -3 -2 2 3

Olegator [25]

Answer:

-10

Step-by-step explanation:

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3 years ago

UDD Question 6 10 pts Solve for x. Round to the nearest tenth (one decimal place). Just type the number, no unit of measurement.

NISA [10]

Answer:

The answer is;

54.4

Step-by-step explanation:

The given parameter are;

Triangle ΔFGH = Right triangle

The length of segment $\overline {FH}$ = 48

The measure of ∠HFG = 28°

The measurement required = The measure of segment $\overline {FG}$ = x

The measure of ∠FHG = 90° angle opposite hypotenuse side of a right triangle

Therefore, ∠FGH = 180° - ∠HFG - ∠FHG = 180° - 90° - 28° = 62°

∠FGH = 62°

By sine rule, we have;

$\overline {FH}$ /(sin ∠FGH) = $\overline {FG}$ /(sin(∠FHG)

By substituting the known values, we have;

48/(sin 62°) = x/(sin(90°)

sin(90°) = 1, therefore, we have;

x/1 = x = 48/(sin 62°) = 54.4 (by rounding the answer to the nearest tenth)

x = 54.4

8 0

2 years ago

At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul

vampirchik [111]

Answer:

a) Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b) The 95% confidence interval would be given by (910.05;959.95)

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) $z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750$

Since is a bilateral test the p value is given by:

$p_v =2*P(Z>2.750)=0.0059$

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b. What is the 95% confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=935$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=180$ represent the population standard deviation

n=200 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=3278.222$

The sample deviation calculated $s=97.054$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$