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3 years ago

Helppp?!!!!!!!!!!!????!!!?????

Mathematics

1 answer:

Shalnov [3]3 years ago

7 0

Answer:

-1.

Step-by-step explanation:

The difference of -3 and a number is -2. Find the number:

-3 - x = -2

x = -2 + (-3)

x = -1

-1 is the answer.

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The cafeteria at Midtown Middle school surveyed 575 students about their favorite food. Determine the number of students that Re

Helen [10]

You would put this into a fraction
so for the salad
$\frac{20}{100} = \frac{?}{575}$
and for the fruit
$\frac{24}{100} = \frac{?}{575}$
cross multiply and
1. salad: 20% = 115 students
2. Fruit: 24% = 138 students

8 0

3 years ago

A random sample of 10 chocolate energy bars of a certain brand has, on average, 230 calories per bar, with a standard deviation

Troyanec [42]

Answer:

The confidence interval for the true mean calorie content is

$215$

Step-by-step explanation:

In this problem we know the mean and standard deviation fo the sample, so we can compute the CI like this:

$\bar{x}-t*s/\sqrt{n}< \mu$

We need to determine t for 10-1=9 degrees of freedom and 99% confidence level. We look up in the t-table and the value for t is 3.2498.

Then we can calculate the limits of the CI:

$\bar{x}-t*s/\sqrt{n}< \mu$

5 0

4 years ago

Translate this sentence into an equation.

yawa3891 [41]

Answer:

g+13=48

Step-by-step explanation:

3 0

3 years ago

A high school runs a survey asking students if they participate in sports. The results are found below. Run an independence test

Sveta_85 [38]

Answer:

It can be concluded that participation in sports is dependent on grade level.

Step-by-step explanation:

In this case a Chi-square independence test for the data is to be performed at α = 0.01.

The hypothesis can be defined as follows:

H₀: The participation in sports is independent of grade level.

Hₐ: The participation in sports is dependent of grade level.

The data provided is:

Freshmen Sophomores Juniors Seniors

Yes 75 88 55 42

No 30 28 38 40

The formula to compute the expected frequencies is:

$E_{i}=\frac{i^{th}\ \text{Row Total}\ \times\ i^{th}\ \text{Column Total}}{N}$

The Chi-square statistic is:

$\chi^{2}=\sum {\frac{(O_{i}-E_{i})^{2}}{E_{i}}}$

Consider the Excel file attached below.

The value of Chi-square statistic is 16.244.

The degrees of freedom of the test are:

$\text{df}=(r-1)(c-1)$

$=(4-1)(2-1)\\=3\times 1\\=3$

Compute the p-value of the test as follows:

$p-value=P(\chi^{2}_{df}$

$=P(\chi^{2}_{3}$

*Use a Chi-square table.

p-value = 0.001 < α = 0.01

The null hypothesis will be rejected at 1% level of significance.

Thus, it can be concluded that participation in sports is dependent on grade level.

4 0

4 years ago

Rose writes down all of her math quiz scores. Her math scores are 98,78,84, and 96. What is the mean of her math quiz scores?

KATRIN_1 [288]

Answer:

Step-by-step explanation:

Hi, there to find the mean

you need add the numbers and count how many numbers are there and divide

In this case 98+78+84+96= 356

Now we will count

there are 4 numbers

$\frac{356}{4}$ = 89

Your mean will be 89

If you ever have trouble with mean remember this

Mean $\frac{SumofDataPoints}{NumberofDataPoints }$

well it is supposed to be:

Sum of Data Points/Number of Data Points

4 0

3 years ago