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Ivenika [448]

In 127 games for the Barons, MJ managed just three home runs, 17 doubles, and one triple. All told, it amounted to a . 266 slugging percentage, which is puny by any measur

5 0

3 years ago

Factorise x squared + 3x + 2

Zielflug [23.3K]

Answer:

$(x + 2)(x + 1)$

Step-by-step explanation:

by factorising ;

=》

${x}^{2} + 3x + 2$

=》

${x}^{2} + 2x + x + 2$

=》

$x(x + 2) + 1(x + 2)$

=》

$(x + 2)(x + 1)$

i hope it helped.....

8 0

2 years ago

What is the y -value of the solution to the system of equations shown? 6x=3y−24 4x+5y=−2

Vlada [557]

Answer:

y = 2 and x = -3 the solutions of the equation

8 0

2 years ago

Birth weights at a local hospital have a Normal distribution with a mean of 110 oz and a standard deviation of 15 oz. The propor

OLga [1]

Answer:

The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 110, \sigma = 0.15$

The proportion of infants with birth weights between 125 oz and 140 oz is

This is the pvalue of Z when X = 140 subtracted by the pvalue of Z when X = 125. So

X = 140

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{140 - 110}{15}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

X = 125

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{125 - 110}{15}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

0.9772 - 0.8413 = 0.1359

The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.

4 0

2 years ago

A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of seco

Brut [27]

Answer:

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

$df=n_{A}+n_{B}-2=26+24-2=48$

Since is a one sided test the p value would be:

$p_v =P(t_{(48)}>1.921)=0.0303$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X_{A}=48250$ represent the mean of elementary teachers

$\bar X_{B}=45630$ represent the mean for secondary teachers

$s_{A}=3900$ represent the sample standard deviation for elementary teacher

$s_{B}=5530$ represent the sample standard deviation for secondary teachers

$n_{A}=26$ sample size selected

$n_{B}=24$ sample size selected

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{B}\leq 0$

Alternative hypothesis: $\mu_{A}-\mu_{B}>0$

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{A}-\bar X_{B})}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{A}+n_{B}-2=26+24-2=48$

Since is a one sided test the p value would be:

$p_v =P(t_{(48)}>1.921)=0.0303$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

5 0

2 years ago