Ok so I messed up the first time I put only 3 pictures when there was more the questions are in the pictures

a basketball court is a rectangle that is 94 feet long the area is 4700 square feet what is the width of the basketball court

andre [41]

Answer:

A=l x w 4700=94x? 4799/94~jz

7 0

3 years ago

Find the area of a square with a side measure of 3x − 6 feet.

Wittaler [7]

Area of a square: s^2

Because we know the length of one side, we can multiply it by itself to find the area.

(3x - 6)(3x - 6)

9x^2 - 18x - 18x + 36

Combine like terms.

9x^2 - 36x + 36

<h3>The correct answer is A, or the first answer. </h3>

6 0

3 years ago

ANSWER QUICK. BRAINLIEST FOR FASTEST AND BEST ANSWER.

WINSTONCH [101]

(2,40) is the ordered pair where the 2 lines meet

4 0

3 years ago

The amount of $15,000 is invested in two funds paying 2% and 5% simple interest. If the annual interest is $540, how much of the

Lostsunrise [7]

Answer:

The amount $15000 is divided as

in 2% = $7000

in 5% = $8000

Step-by-step explanation:

It is given that the total amount is $15000 and rates of interests are 2% and 5%.

<em>Let the amount invested in 2% be "x", so the amount invested in 5% is (15000 - x).</em>

The annual interest on x part is given as,

= $(\frac{2}{100})(x)$

Similarly, The annual interest on (15000 - x) part is given as,

= $(\frac{5}{100})(15000 - x)$

the total annual interest is given as $540

Thus, <em>540 = $(\frac{5}{100})(15000 - x) + (\frac{2}{100})(x)$ </em>

x = $\frac{21000}{3}$ = 7000

Thus the shares are $7000 and (15000 - 7000) $8000 .

8 0

3 years ago

We are conducting a hypothesis test to determine if fewer than 80% of ST 311 Hybrid students complete all modules before class.

Juli2301 [7.4K]

Answer:

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

Null hypothesis: $p\geq 0.8$

Alternative hypothesis: $p < 0.8$

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

$p_v =P(Z$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

$\hat p=\frac{97}{110}=0.882$ estimated proportion of students who completed the modules before class

$p_o=0.8$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) 2) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%: Null hypothesis:[tex]p\geq 0.8$

Alternative hypothesis: $p < 0.8$

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:

$z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124$

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

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

$p_v =P(Z$

So the p value obtained was a very high value and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

$p_v =P(Z>2.124)=0.013$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

5 0

3 years ago