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2 years ago

12

Joanie runs the concession stand for the school's baseball games. At each game, the best selling items are pretzels and hot dogs

. Pretzels are $3 each, and hot dogs are $2 each. Today she sold 15 more hot dogs than pretzels and made $195 in total sales.
Use a system of equations to model the situation above, and determine which of the following are possible amounts of pretzels and hot dogs that Joanie sold today.

A.
35 pretzels, 45 hot dogs
B.
15 pretzels, 30 hot dogs
C.
33 pretzels, 48 hot dogs
D.
31 pretzels, 51 hot dogs

1 answer:

STALIN [3.7K]2 years ago

3 0

77 because add them all up then subtract 14

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2 years 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:

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

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2 years ago

24=7/8g I need help with this equation and explain how to do it please?

Gnesinka [82]

<span> add</span><span> up the two equivalent fractions </span>
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3 years ago

Suppose your statistics professor reports test grades as z-scores, and you got a score of 2.49 on an exam. (a) Write a sentenc

Oduvanchick [21]

Answer:

a) A z-score of 2.49 means that your grade was 2.49 standard deviations above the mean of all grades.

b) 1.39% of the class scored lower than your friend.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

You got a score of 2.49 on an exam.

(a) Write a sentence explaining what that means

A z-score of 2.49 means that your grade was 2.49 standard deviations above the mean of all grades.

(b) Your friend got a z-score of negative 2−2. If the grades satisfy the Nearly Normal Condition, about what percent of the class scored lower than your friend? 

This percentage is the pvalue of $Z = -2.2$

So 0.0139 = 1.39% of the class scored lower than your friend.

7 0

3 years ago

There are 2504 computer science students at a school. Of these, 1876 have taken a course in Java, 999 have taken a course in Lin

Y_Kistochka [10]

Answer:

492

Step-by-step explanation:

As per the given data of the question:

Total number of students = 2504

Number of students in Java (J) = 1876

Number of students in Linux (L) = 999

Number of students in C = 345

J∩L = 876

L∩C = 231

C∩J = 290

L∩J∩C = 189

Now according to Venn-diagram as drawn below:

Number of students haven taken courses in Java and Linux both only

= J∩L - L∩J∩C

= 876 - 189

= 687

Number of students haven taken courses in Java and C both only

= C∩J - L∩J∩C

= 290 - 189

= 101

Number of students haven taken courses in C and Linux both only

= L∩C - L∩J∩C

= 231 - 189

= 42

Therefore,

Number of students only in Java = 1876 - 687 - 189 - 101 = 899

Number of students only in Linux = 999 - 687 - 189 - 42 = 81

Number of students only in C = 345 - 42 - 189 - 101 = 13

So,

Number of students who have not taken a course in any of these three subjects

= 2504 - 899 - 81 - 13 - 687 - 189 -101 - 42

= 492

Hence, the students who have not taken a course in any of these three subjects = 492.

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2 years ago