Arrange the functions in increasing order of their periods y=-3tan(x+2

Which is an example of a quadratic function whose graph is entirely above the x-axis?

vredina [299]

The answer is: [D]: " y = x² + 2x + 2 " .
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7 0

4 years ago

Carlos is tiling a kitchen cluster that is 12 feed by 3 feed. The counter has a rectangular hole 3 feet by 2 feet cut in ir for

Taya2010 [7]

Answer:

The area of the counter that Carlos will wall tile is $30 ft^{2}$

Step-by-step explanation:

to solve this, we will calculate the area of the counter and the area of the rectangular hole that Carlos would have to leave out for the sink, and then subtract them to get the area that will be tiled.

Total area of the rectangular counter = length X breadth = $12 \times 3 = 36ft^{2}$

area of the rectangular cut out for the sink = length X breadth = $3 \times 2 = 6 ft^{2}$

Surface area that Carlos is meant to tile = $36ft^{2}- 6 ft^{2} = 30ft^{2}$

Hence,the area of the counter that Carlos will wall tile is $30 ft^{2}$

6 0

3 years ago

A teacher of statistics wants to know if a new teaching methodology that includes IT is efficient in terms of increased average

iragen [17]

Answer:

The answer is 0.31

Step-by-step explanation:

<u>Old Method New Method .</u>

Mean 73.5625 Mean 75.70588

Standard Error 3.143736 Standard Error 2.923994

Median 72 Median 75

Mode 90 Mode 64

Standard deviation 12.57494 Standard deviation 12.05594

Sample Variance 158.1292 Sample Variance 145.3456

Kurtosis -1.14544 Kurtosis -0.76646

Skewness 0.171025 Skewness 0.091008

Range 39 Range 41

Minimum 55 Minimum 56

Maximum 94 Maximum 97

Sum 1177 Sum 1287

Count 16 Count 17

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μNew< μOld

Alternative hypothesis: μNew > μOld

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

$SE=\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } \\\\SE=4.29$

DF = 31

$t = \frac{(x_1-x_2)-d}{SE} \\\\t = - 0.4997$

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of - 0.499. We use the t Distribution Calculator to find P(t < - 0.499) = 0.311

Therefore, the P-value in this analysis is 0.311.

Interpret results. Since the P-value (0.311) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do have sufficient evidence in the favor of the claim that new method is efficient than the old method.