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2 years ago

13

During the 2005 football season, Team Tackle beat Team Sack by 3 points. If their combined scores totaled 15, find the individ

ual team scores. Team Tackle's score was ? points. Team Sack's score was ? points.

2 answers:

Verizon [17]2 years ago

7 0

Answer:

Team tackle scored 9 points.

Team Sack scored 6 points.

Step-by-step explanation:

9+6=15 and 9 is three more than 6.

Igoryamba2 years ago

3 0

Team sack = 9
Team tackle = 6

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Mitchel owns a livestock trailer that can hold a maximum of 5,000 pounds. The average weight of each goat is 90 pounds, and the

castortr0y [4]

Answer:

The number of cows and calves Mitchel can transport is determined by the inequity $90g+360c \leq 5000$

Step-by-step explanation:

Let $g$ be the number of goats, and $c$ be the number of cows. If Mitchel's livestock trailer can only hold maximum of 5000 pounds. then we have the inequality

$90g+360c \leq 5000$ <em>(this says that the weight of the goats and the calves cannot exceed 500 pounds.) </em>

Therefore, this inequality determines the number of goats and calves Mitchel can take in a single trip.

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2 years ago

After the 2008 elections, it is desired to estimate the proportion of Florida voters who now regret that they did not vote.

Sonja [21]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded we got 1681

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".

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.10$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.02$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimates proportion is 0.5 since we don't have other info provided to assume a different value. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded we got 1681

8 0

2 years ago

A sample of bacteria is being eradicated by an experimental procedure. The population is following a pattern of exponential deca

77julia77 [94]

Answer:

There will be 50 bacteria remaining after 28 minutes.

Step-by-step explanation:

The exponential decay equation is

$N=N_0e^{-rt}$

N= Number of bacteria after t minutes.

$N_0$ = Initial number of bacteria when t=0.

r= Rate of decay per minute

t= time is in minute.

The sample begins with 500 bacteria and after 11 minutes there are 200 bacteria.

N=200

$N_0$ = 500

t=11 minutes

r=?

$N=N_0e^{-rt}$

$\therefore 200=500e^{-11r}$

$\Rightarrow e^{-11r}=\frac{200}{500}$

Taking ln both sides

$\Rightarrow ln| e^{-11r}|=ln|\frac{2}{5}|$

$\Rightarrow {-11r}=ln|\frac{2}{5}|$

$\Rightarrow r}=\frac{ln|\frac{2}{5}|}{-11}$

To find the time when there will be 50 bacteria remaining, we plug N=50, $N_0$ = 500 and $r}=\frac{ln|\frac{2}{5}|}{-11}$ in exponential decay equation.

$50=500e^{-\frac{ln|\frac25|}{-11}.t}$

$\Rightarrow \frac{50}{500}=e^{\frac{ln|\frac25|}{11}.t}$

Taking ln both sides

$\Rightarrow ln|\frac{50}{500}|=ln|e^{\frac{ln|\frac25|}{11}.t}|$

$\Rightarrow ln|\frac{1}{10}|={\frac{ln|\frac25|}{11}.t}$

$\Rightarrow t= \frac{ln|\frac{1}{10}|}{\frac{ln|\frac25|}{11}.}$

$\Rightarrow t= \frac{11\times ln|\frac{1}{10}|}{{ln|\frac25|}}$

$\Rightarrow t\approx 28$ minutes

There will be 50 bacteria remaining after 28 minutes.

3 0

2 years ago