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Mathematics

1 answer:

AlladinOne [14]3 years ago

8 0

Answer:

These relationships are realted because the variable is multiplied by ten to get the function.

9cm: 90mm

12 1/4cm: 122.5mm

50cm: 500mm

88.49cm: 884.9mm

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A shoe manufacturer collected data regarding men's shoe sizes and found that the distribution of sizes exactly fits the normal c

Lina20 [59]

Answer: 97.72%

Step-by-step explanation:

Given : A shoe manufacturer collected data regarding men's shoe sizes and found that the distribution of sizes exactly fits the normal curve.

Let x be the random variable that represents the shoe sizees.

Also, The population mean = $\mu=11$ ; Standard deviation: $\sigma=1.5$

Formula for z:-

$z=\dfrac{x-\mu}{\sigma}$

Put x= 8, we get

$z=\dfrac{8-11}{1.5}=-2$

Now, the probability that the male shoe sizes are greater than 8 :-

$P(x>8)=P(z>-2)=1-P(z\leq-2)\\\\=1-0.0227501=0.9772499\approx0.9772$

Hence, the percent of male shoe sizes are greater than 8 is 97.72%.

7 0

3 years ago

As the carbon content in steel increases, its ductility tends to decrease. A researcher at a steel company measures carbon conte

mixer [17]

Answer:

d. -5.666 to -0.926

Step-by-step explanation:

Here a pivotal quantity is $T = \frac{\hat{\beta_{1}}-\beta_{1}}{se(\hat{\beta_{1}})}$ where $\beta_{1}$ is the true slope of the regression equation (unknown), $\hat{\beta_{1}}$ is its least square estimate and $se(\hat{\beta_{1}})=1.097$ is its estimated standard error. And T has t-distribution with n-2=15-2=13 degrees of freedom (n is the sample size). We have that $\hat{\beta_{1}} = -3.30$ , and because we want the 95% confidence interval, we should use the 2.5th quantile of the t distribution with 13 df, this value is -2.16 and the 95% confidence intervale is given by $\hat{\beta_{1}}\pm t_{0.025}se(\hat{\beta_{1}})$ , i.e., $-3.30\pm (2.16)(1.097)$ . Therefore, the 95% confidence interval explicitly is (-5.666, -0.926)