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bogdanovich [222]
3 years ago
10

HELPPPP PLEASEEE!!! I NEED HELPP

Mathematics
1 answer:
zhenek [66]3 years ago
6 0

Answer:

≥≤ are used when the number is less/greater or equal than some thing.

>< are used when a number is strictly less/greater than some thing.

1) a number x is less than 0 or at least 3

A number x is less than 0 is written as: x < 0.

x is at least 3 is written as: x ≥ 3.

Then we can write the statement as:

x < 0  ∨ x ≥ 3.

where ∨ means "or"

2) a number x is less than or equal to 4 and greater than -1

x less than or equal to 4 is written as: x ≤ 4

x is greater than -1 is written as: x > - 1.

We can write the statement as:

-1 < x ≤ 4

3) a number x is less than 8 and greater than or equal to 5

x is less than 8 is written as: x < 8

x is greater than or equal to 5 is written as: x ≥ 5.

Then the statement can be written as:

5 ≤ x < 8,

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s2008m [1.1K]

Answer:

y = x - 10

Step-by-step explanation:

3 0
3 years ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \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 for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

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{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0
3 years ago
Simplify<br> -4 1/5 (-13 1/10)<br> pls help
oksian1 [2.3K]

Answer:

55 1/50 or 2751/50

Step-by-step explanation:

Convert both expressions to improper form

Make their denominators the same

Multiply the numerators

Simplify the fraction if needed

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3 years ago
In a recent survey of American teenagers, 44% of them chose pizza
EastWind [94]

Answer: i dont know the answer

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Which statements are true? *<br> 1:2 = 50:100<br> 2:1 = 50:100<br> 1:2 = 2:1<br> 3:4 = 9:12
Bumek [7]
1:2 = 2:1 is the statement
6 0
3 years ago
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