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4 years ago

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joan counted that 2/10 of her jelly beans were red. Dean counted that 6/10 of his jelly beans were red. Hiw much greater a fract

ion od Dean's jelly beans were red?

1 answer:

ch4aika [34]4 years ago

3 0

If you take the 6 and subtract it by 2 you will get 4

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if matt has 20 apples and then if he gets 121 more of his friends how many apples would he get in total?

mario62 [17]

Answer:

141 would be your answer

Step-by-step explanation:

20+121=141

4 0

3 years ago

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The combined math and verbal scores for females taking the SAT-I test are normally distributed with a mean of 900 and a standard

Genrish500 [490]

Answer:

50% of females do not satisfy that requirement

Step-by-step explanation:

Problems of normally distributed samples are solved using 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 problem, we have that:

$\mu = 900, \sigma = 200$

If a college includes a minimum score of 900 among its requirements, what percentage of females do not satisfy that requirement

This is scores lower than 900, which is the pvalue of Z when X = 900.

So

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

$Z = \frac{900 - 900}{200}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

50% of females do not satisfy that requirement

7 0

3 years ago

Suppose μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of bra

Stella [2.4K]

Answer:

The 95% confidence interval would be given by $-23.44 \leq \mu_1 -\mu_2 \leq -8.16$

Step-by-step explanation:

Previous concepts

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".

$\bar X_1 =114.1$ represent the sample mean 1

$\bar X_2 =129.9$ represent the sample mean 2

n1=5 represent the sample 1 size

n2=2 represent the sample 2 size

$s_1 =5.08$ sample standard deviation for sample 1

$s_2 =5.37$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =114.1-129.9=-15.8$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_1 +n_2 -1=5+5-2=8$

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: "=-T.INV(0.025,8)".And we see that $t_{\alpha/2}=\pm 2.31$

The standard error is given by the following formula:

$SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=3.306$

Confidence interval

Now we have everything in order to replace into formula (1):

$-15.8-2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-23.437$

$-15.8+2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-8.163$

So on this case the 95% confidence interval would be given by $-23.44 \leq \mu_1 -\mu_2 \leq -8.16$

8 0

3 years ago

Find the limit<br> Picture below

pogonyaev

Answer:

a. 4

Step-by-step explanation:

We want to find;

$\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$

If $f(x)=x^2$ .

$f(2+h)=(2+h)^2$

$f(2+h)=4+4h+h^2$

Also;

$f(2)=2^2$

$f(2)=4$

This implies that;

$\lim_{h \to 0} \frac{4+4h+h^2-4}{h}$

Simplify;

$\lim_{h \to 0} \frac{4h+h^2}{h}$

Factor to get;

$\lim_{h \to 0} \frac{h(4+h)}{h}$

Cancel the common factors.

$\lim_{h \to 0} 4+h$

Substitute h=0 to get

$\lim_{h \to 0} 4+h=4+0=4$

7 0

4 years ago

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What is the slope of the line graphed below

bonufazy [111]

1/5 the slop is the rise over run

3 0

3 years ago

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