PLEASE GUYS HELP ME !!!!!

Use ΔDEF, shown below, to answer the question that follows:

ddd [48]

Answer: 41.51

Step-by-step explanation:

$\sin 49^{\circ}=\frac{x}{55}\\x=55 \sin 49^{\circ} \approx \boxed{41.51}$

8 0

2 years ago

PLZ HELP I WILL MAKE YOU BRAINLIEST!! IT IS GOOD FOR YOUR PROFILE!

ella [17]

1 11/25 = 36/25
= (6/5)^2

3 0

3 years ago

Read 2 more answers

k.jennifer went to the mall with $290 she spent$115 on clothes $75 on new shoes how muchdoes Jennifer have left

stira [4]

Answer:

$100

Step-by-step explanation:

First add up how much she spent

$115+75= $190

Then Subtract from how much she had to spend

$290-190= $100

hope this helps:)

6 0

3 years ago

Read 2 more answers

Bill has 205 in his savings account and adds 5$ per week.John has $5 in his account and adds 25 dollers per week. After six week

stiks02 [169]

Answer:

bill

Step-by-step explanation:

Bill:

205+(6x5)

205+30= 235

Bill - $235

John:

5+(25x6)

5+150=155

John - $155

So bill has more

Hope this helps

8 0

3 years ago

A machine in the student lounge dispenses coffee. The average cup of coffee is supposed to contain 7.0 ounces. A random sample o

Elenna [48]

Answer:

$t=\frac{7.4-7}{\frac{0.7}{\sqrt{7}}}=1.512$

The degrees of freedom are

$df=n-1=7-1=6$

And the p value for this case is:

$p_v =2*P(t_{(6)}>1.512)=0.181$

The p value is a higher value and using a significance levels of 5% or 10% we see that the p value is higher than the significance level and then we FAIL to reject the null hypothesis and we don't have enough evidence to conclude that the true mean is different from 7 ounces.

Step-by-step explanation:

Information provided

$\bar X=7.4$ represent the sample mean

$s=0.7$ represent the sample standard deviation

$n=7$ sample size

$\mu_o 7$ represent the value to verify

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the average amount of coffee per cup is different from 7 ounces, the system of hypothesis would be:

Null hypothesis: $\mu = 7$

Alternative hypothesis: $\mu \neq 7$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{7.4-7}{\frac{0.7}{\sqrt{7}}}=1.512$

The degrees of freedom are

$df=n-1=7-1=6$

And the p value for this case is:

$p_v =2*P(t_{(6)}>1.512)=0.181$

The p value is a higher value and using a significance levels of 5% or 10% we see that the p value is higher than the significance level and then we FAIL to reject the null hypothesis and we don't have enough evidence to conclude that the true mean is different from 7 ounces.

7 0

3 years ago