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3 years ago

I need help fast……….

Mathematics

1 answer:

kap26 [50]3 years ago

8 0

Answer:

all real numbers

the domain of any parabola is all real numbers because it gets wider.

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76 is 80% of what number

Shkiper50 [21]

Answer:

Step-by-step explanation:

6 0

3 years ago

On the same day at school, 1/6 of the students were wearing skirts and 5/8 of the students were wearing pants. The rest of the s

anastassius [24]

What you do is you find the common denominator between the two fractions given. the easiest is to just multiply them

6*8=48

and then multiply each numerator by the number it's denominator is multiplied by

1*8=8 and 5*6=30

and now you have your new fractions, add them together

8/48+30/48=38/48

this leaves 10/48 as the people who were wearing shorts, just simplify this and it's done

10/2 =5/24
48/2

4 0

3 years ago

Read 2 more answers

What is the simplified form of this expression? (3 - 13x - 7x?) - (5X2 + 12x - 10)

tatiyna

Answer:

-5x^2-32x+13

Step-by-step explanation:

(3-13x-7x)-(5x^2+12x-10)

3-20x-5x^2-12x+10

-5x^2-32x+13

6 0

3 years ago

Read 2 more answers

A car salesman sells cars with prices ranging from $5,000 to $45,000. The histogram shows the distribution of the numbers of car

Ierofanga [76]

Answer:

D. The mean will shift to the left.

Because if they sell lower priced cars it will bring the average price down causing the mean to move to the left

7 0

3 years ago

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A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the stor

mixer [17]

Answer:

Critical value is t = 1.9901

Step-by-step explanation:

We are given the results of the sampling :

In-House Credit Card National Credit Card Sample Size : 32 50

Mean Monthly Purchases : $45.67 $39.87

Standard Deviation : $10.90 $12.47

Also, the managers wished to test whether there is a statistical difference in the mean monthly purchases by customers using the two types of credit cards, using a significance level α of 0.05.

Firstly, we will specify our null and alternate hypothesis;

Let $\mu_1$ = Mean Monthly Purchases of In-House Credit Card

$\mu_2$ = Mean Monthly purchases of National Credit Card

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in the mean monthly purchases by customers using the two types of credit cards}

Alternate Hypothesis, $H_0$ : $\mu_1-\mu_2\neq$ 0 {means that there is statistical difference in the mean monthly purchases by customers using the two types of credit cards}

The test statistics that will be used here is Two-sample t-test statistics;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n___1+n_2-_2$

where, $\bar X_1$ = Sample mean Purchases of In-House Credit Card = $45.67

$\bar X_2$ = Sample mean Purchases of National Credit Card = $39.87

$s_p$ = pooled variance

$n_1$ = In-house credit card sample = 32

$n_2$ = National credit card sample = 50

So, degree of freedom of t-value here is (32 + 50 - 2) = 80

Now, at 0.05 significance level, t table gives critical value of t = 1.9901 at 80 degree of freedom.

Therefore, the critical value assuming the population standard deviations are not known but that the populations are normally distributed with equal variances is t = 1.9901.

4 0

3 years ago