What is the total surface area of the cone? 90π cm2 65π cm2 120π cm2 115π cm2

Can you guyz help me with this question.please write the answers step by step ☺️

sasho [114]

Hey! Ok so with this problem, you can figure out the pattern with subtraction.

If you convert the fractions to decimals you get 3.25 and 5.625.
5.625 - 3.25 = 2.375, which means he increased his run by 2.375 miles.
8 - 5.625 = 2.375 as well, which follows the pattern. So what you would need to do from here is add 2.375 miles for each day he jogged.

8 + 2.375 = 10. 375, the miles he jogged on thursday.
10.375 + 2.375 = 12. 75, the miles he jogged on friday.

So, when converted to fractions, your answer is D: Eric jogged 12 3/4 miles on Firday

7 0

3 years ago

72times45 Help Please

worty [1.4K]

Answer:

3,240

Step-by-step explanation:

Multiply 72 x 45

7 0

3 years ago

Read 2 more answers

Sam hiked 4.5 km along a straight bush track, and then turned 40° and walked a further 3.8 km in a straight line.

erica [24]

Answer is 8.3 hope you is happy

3 0

1 year ago

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

Jane buys some rocks for decorating her backyard. She uses a box full of rocks. The box is a rectangular container that is 1 yar

statuscvo [17]

First, Jane must know the volume of the box.
Volume = l * w * h
Volume = 1 * 0.5 * 0.5
Volume = 0.25 yard^3

1 yard^3 = $25

So,
1 : 25 = 0.25 : x
x = 25 * 0.25
x = $6.25

So she need to spend $6.25

4 0

3 years ago