6. Write and solve a system of THREE Inequalities that models the situation below.

Harry has 20 sweets. He gives 7 of the sweets to Nadia. What fraction of the 20 sweets does Harry have now?

olganol [36]

Answer: 13

Step-by-step explanation:

1. Harry has 20/20 sweets

2. Nadia takes 7/20 sweets

3. Your left with 13/20

3 0

4 years ago

Read 2 more answers

A radiator contains 15 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure anti

Annette [7]

%Antifreeze=VAntifreezeVFluid VFluid=VAntifreeze%Antifreeze VAntifreeze=Vfluid∗%Antifreeze I want to find the amount of antifreeze in a 15 quart solution with 30% antifreeze VAntifreeze=15∗0.30 =18/4 quarts of antifreeze Similarly, I want to find the amount of antifreeze in a 15 quart solution with 35% antifreeze first. VAntifreeze=15∗0.35 = 21/4 quarts of antifreeze

the difference between 21/4 and 18/4 is 3/4 quarts, which is the amount of pure antifreeze I've added in.

SO the V_fluid I replaced with 3/4 quarts of antifreeze is (3/4)/ 0.35

5 0

3 years ago

I need help bad I will mark you as king or what every pkz get it right I can’t get it right plz help

Ostrovityanka [42]

Answer:

d=6c it looks like. good luck

4 0

3 years ago

This circle, with center point Q, has a radius of 10 centimeters. The length of the minor arc NP is 20.42 centimeters. To the ne

maxonik [38]

The complete question in the attached figure

we know that
[the length of a circumference]=2*pi*r
for r=10 cm
[the length of a circumference]=2*pi*10-----> 62.8 cm

if 360° ( full circle) has a length of---------> 62.8 cm
x--------------------------------------> 20.42 cm
X=20.42*360/62.8-----> x+117.05°-------> x=117°

the answer is
the value of x is 117°

6 0

3 years ago

Use this information to answer the questions. University personnel are concerned about the sleeping habits of students and the n

Oksanka [162]

Answer:

$z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097$

$p_v =P(Z>2.097)=0.018$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Step-by-step explanation:

1) Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

$\hat p=\frac{209}{377}=0.554$ estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p > 0.5$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(Z>2.097)=0.018$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

6 0

3 years ago