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3 years ago

6

In response to very high demand, a basketball jersey has been marked up by 25% and is now selling for $95.00. How much was the j

ersey before the mark-up?

1 answer:

Fynjy0 [20]3 years ago

6 0

I believe it would be 71.25%.

Hope this helps

Have a nice day.

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Please help giving brainlest

Likurg_2 [28]

Answer:

8 ft; 6 m; 28pi ft; 10pi mm

Step-by-step explanation:

1. Radius = diameter/2

16/2=8

Therefore, 8 ft

2. Diameter= radius * 2

3*2=6

Therefore, 6 m

3. Circumference = 2*pi*r

2*pi*14=28pi

Therefore, 28pi ft

4. Circumference = pi*d

Therefore, 10pi mm

I hope this helped and have a good rest of your day!

6 0

2 years ago

Read 2 more answers

The following data comparing wait times at two rides at Disney are listed below: Position Pirates Splash Mountain Sample Size 32

myrzilka [38]

Answer:

a) $(14.68 -18.77) - 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =-12.968$

$(14.68 -18.77) + 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =4.788$

b) $t=\frac{14.68-18.77}{\sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}}}}=-1.10$

Step-by-step explanation:

Data given and notation

$\bar X_{A}=14.68$ represent the mean for Pirates

$\bar X_{B}=18.77$ represent the mean for Splash Mountain

$s_{A}=11.87$ represent the sample standard deviation for the sample Pirates

$s_{B}=16.79$ represent the sample standard deviation for the sample Slpash Mountain

$n_{A}=32$ sample size selected for Pirates

$n_{B}=30$ sample size selected for Splash Mountain

$\alpha=0.02$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

Part a

The confidence interval would be given by:

$(\bar X_A -\bar X_B) \pm t_{\alpha/2} \sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}$

The degrees of freedom are given by:

$df = n_A +n_B -2 = 32+30-2 = 60$

Since we want 98% of confidence the significance level is $\alpha =1-0.98 =0.02$ and $\alpha/2 =0.01$ , we can find in the t distribution with df =60 a critical value that accumulates 0.01 of the area on each tail and we got:

$t_{\alpha/2}= 2.39$

And replacing we got for the confidence interval:

$(14.68 -18.77) - 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =-12.968$

$(14.68 -18.77) + 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =4.788$

Part b

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the means are equal, the system of hypothesis would be:

Null hypothesis: $\mu_{A} = \mu_{B}$

Alternative hypothesis: $\mu_{A} \neq \mu_{B}$

the statistic is given by:

$t=\frac{\bar X_{A}-\bar X_{B}}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{14.68-18.77}{\sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}}}}=-1.10$

6 0

2 years ago

Awnser These Do not just right What or something or i will report!

Brrunno [24]

1. 37
2. 5
3. 10
4. 4
5. -3

3 0

2 years ago

Read 2 more answers

Write an expression equivalent to h^6/h^3 in the form of h^m

Pavel [41]

Answer:

$h^3 = h^m$

Step-by-step explanation:

$\frac{h^6}{h^3} = h^m$

6 - 3 = 3

keep the base same, therefore

$h^3 = h^m$

Hope this helps!

3 0

3 years ago

Multiply. 1 1/2 x 3 2/3

sasho [114]

Answer:

5.5

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers