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VARVARA [1.3K]
3 years ago
7

June serve 4/9 of the fruit punch . how many litters does she have left?

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
3 0
18- 4/9 = 17 9/9 - 4/9 = 17 5/9 litters
You might be interested in
10 points.
rusak2 [61]

Answer:

<u>The difference between the experimental probability and the theoretical probability is 0.08 (rounding the answer to the nearest hundredth) or 8%.</u>

Correct statement and question:

Regina has a bag of 6 orange marbles and 6 black marbles. She picks a marble at random and then puts it back in the bag. She does this 24 times. The results can be found in the table.

Outcome Tally

Orange  10

Black     14

Figure out the percent error of pulling a black marble in Regina’s experiment. Show your work and round the answer to the nearest hundredth.

Source:

Previous question that can be found at brainly

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Number of orange marbles = 6

Number of black marbles = 6

Number of times of the experiment = 24

Number of times the outcome was an orange marble = 10

Number of times the outcome was a black marble  = 14

2. Figure out the percent error of pulling a black marble in Regina’s experiment. Show your work and round the answer to the nearest hundredth.

The theoretical probability of pulling a black marble is 12/24 or 0.5, given that the number of orange and black marbles are equal and actually, the experimental probability is 14/24 or 0.5833.

We can't describe this difference as an "error". What happened here is that there is a difference between the experimental probability and the theoretical probability.

0.5833 - 0.5 = 0.0833 = 0.08 (rounding the answer to the nearest hundredth)

<u>The difference between the experimental probability and the theoretical probability is 0.08 (rounding the answer to the nearest hundredth) or 8%.</u>

5 0
3 years ago
3 1/8 ÷(x−4 7/28 )= 17/18 +1 5/6
sp2606 [1]
Welcome :)
==========

4 0
3 years ago
Read 2 more answers
En una muestra de 50 restaurantes de comida rápida, la venta media fue de $3.000, y la desviación estándar, $200. El intervalo d
Step2247 [10]

Respuesta:

(2945.411; 3054.589)

Explicación paso a paso:

Dado ;

Tamaño de la muestra, n = 50

Media, xbar = 3000

Desviación estándar, s = 200

Nivel de confianza, Zcrítico al 95% = 1,96

El intervalo de confianza se define como:

Xbar ± margen de error

Margen de error = Zcrítico * s / sqrt (n)

Margen de error = 1,96 * 200 / sqrt (50)

Margen de error = 54.589

Límite inferior = (3000 - 54.589) = 2945.411

Límite superior = (3000 + 54.589) = 3054.589

(2945.411; 3054.589)

8 0
3 years ago
Deminsions of a swedish flag in inches
Tanya [424]

Answer: 24.75 x 1.5 x 1.5 inches

Step-by-step explanation: Hope this helps!

8 0
3 years ago
A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores
EastWind [94]

Answer:

(52-49) -4.300= -1.300

(52-49) +4.300= 7.300

And the 95% confidence would be :

-1.300 \leq \mu_1 -\mu_2 \leq 7.300

Step-by-step explanation:

We have the following info given from the problem

\bar X_1 = 52 sample mean for this year

s_1= 13 sample deviation for this year

n_1 = 75 random sample selected for this year

\bar X_2 = 49 sample mean for last year

s_2= 11 sample deviation for last year

n_1 = 53 random sample selected for last year

And we want to construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales

For this case the formula that we need to use is:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom are given by:

df= n_1 +n_2 -2 = 75+53-2= 126

The confidence level is 0.95 and the significance would be \alpha=0.05 and \alpha/2 =0.025 so then the critical value for this case is :

t_{\alpha/2}= 1.979

The margin of error would be:

ME = 1.979 \sqrt{\frac{13^2}{75} +\frac{11^2}{49}}= 4.300

And the confidence interval would be given by:

(52-49) -4.300= -1.300

(52-49) +4.300= 7.300

And the 95% confidence would be :

-1.300 \leq \mu_1 -\mu_2 \leq 7.300

6 0
3 years ago
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