5+5-6+10-100 what is the answer

Zoe used 3/8 of her ribbon to wrap a present. If she uses 9 feet of ribbon for the present how much ribbon did Zoe have it first

MArishka [77]

Let the total ribbon Zoe had was x feet.

Out of x feet, she used 3/8 part of it.

Hence, $\frac{3x}{8}$

3/8 part of the original ribbon is equal to 9 feet.

Hence, we get

$\frac{3x}{8} = 9$

=> Multiply 8 on both sides.

$3x=72$

Divide both sides by 3

$\frac{3x}{3}=\frac{72}{3}$

$x=24 ft$

Zoe had 24 feet of ribbon in starting... Answer

8 0

3 years ago

HELP ASAP 10 POINTS

Elodia [21]

Answer:

596.223

Step-by-step explanation:

3 0

3 years ago

consuela earns a salary of $40,000 per year plus a commission of $1000 for each car she sells right and solve an equation that s

musickatia [10]

The equasion she would use would be 60,000 = 1000x + 40,000

this would means 1,000 times a number plus the 40,000 shes already getting equals 60,000

7 0

3 years ago

Read 2 more answers

I need help with #17 some body pls help this is due in a few min and it’s my last question T^T

Ilia_Sergeevich [38]

Step-by-step explanation:

i think this would be the right answer...

6 0

3 years ago

Read 2 more answers

Two statistics teachers both believe that each has the smarter class. To put this to the test, they give the same final exam to

SVETLANKA909090 [29]

Answer:

We conclude that there is no difference between the two classes.

Step-by-step explanation:

We are given that two statistics teachers both believe that each has a smarter class.

A summary of the class sizes, class means, and standard deviations is given below:n1 = 47, x-bar1 = 84.4, s1 = 18n2 = 50, x-bar2 = 82.9, s2 = 17

Let $\mu_1$ = mean age of student cars.

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1=\mu_2$ {means that there is no difference in the two classes}

Alternate Hypothesis, $H_A$ : $\mu_1\neq \mu_2$ {means that there is a difference in the two classes}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(47-1)\times 18^{2}+(50-1)\times 17^{2} }{47+50-2} }$ = 17.491

So, the test statistics = $\frac{(84.4-82.9)-(0)}{17.491 \times \sqrt{\frac{1}{47}+\frac{1}{50} } }$ ~ $t_9_5$

= 0.422

The value of t-test statistics is 0.422.

Now, the P-value of the test statistics is given by;

P-value = P( $t_9_5$ > 0.422) = From the t table it is clear that the P-value will lie somewhere between 40% and 30%.

Since the P-value of our test statistics is way more than the level of significance of 0.04, so we have insufficient evidence to reject our null hypothesis as our test statistics will not fall in the rejection region.

Therefore, we conclude that there is no difference between the two classes.

6 0

3 years ago