Which of the following is equivalent to “the product of 2 and 3 is increased by 7”?

a chef uses 4 3/4 cups of broth for 10 servings of soup. How much broth is used in one serving of soup?

mafiozo [28]

Set up a proportion. 4 3/4 cup broth is to 10 servings as x cups broth is to 1 serving soup.

x= amount of broth in one serving

(4 3/4)/10= x/1
cross multiply

(4 3/4)(1)= 10x
convert 4 3/4 to improper fraction

(19/4)(1)= 10x

19/4= 10x
divide both sides by 10

19/4 ÷ 10= x
to divide fractions, multiply by reciprocal of 10

19/4 * 1/10= x
multiply numerators; multiply denominators

(19*1)/(4*10)= x

19/40 cup= x

ANSWER: 19/40 cup in one serving

Hope this helps! :)

4 0

3 years ago

Colton has a 1/20 probability of winning the race. Logan has a 15% probability of winning. Who has the greater chance of winning

LiRa [457]

Answer:

Logan has the greater probability of winning

Step-by-step explanation:

Colton=1/20

=0.05

So the probability of Colton is 0.05

Logan=15%

=15/100

=0.15

So the probability of Logan is 0.15

When comparing the two,it was observed that the Logan has greater probability of winning

4 0

3 years ago

In a random sample of 150 customers of a high-speed Internetprovider, 63 said that their service had been interrupted one ormore

erastovalidia [21]

Answer:

a) The 95% confidence interval would be given by (0.341;0.499)

b) The 99% confidence interval would be given by (0.316;0.524)

c) n=335

d)n=649

Step-by-step explanation:

1) Notation and definitions

$X_{IS}=63$ number of high speed internet users that had been interrupted one or more times in the past month.

$n=150$ random sample taken

$\hat p_{IS}=\frac{63}{150}=0.42$ estimated proportion of high speed internet users that had been interrupted one or more times in the past month.

$p_{IS}$ true population proportion of high speed internet users that had been interrupted one or more times in the past month.

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".

The population proportion have the following distribution

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

1) Part a

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 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

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

If we replace the values obtained we got:

$0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499$

The 95% confidence interval would be given by (0.341;0.499)

2) Part b

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 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: