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Serhud [2]
3 years ago
5

Please Help Thank You I will reward brainliest

Mathematics
2 answers:
jeyben [28]3 years ago
5 0
The scale of a drawing is the ratio that all parts of an object are reduced or magnified by.
Goryan [66]3 years ago
4 0
The scale of a drawing is the RATIO that all parts of an object are reduced or magnified by
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gtnhenbr [62]

Answer:

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Step-by-step explanation:

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8 0
2 years ago
Please help me please
oksano4ka [1.4K]

Answer:

See below.

Step-by-step explanation:

1. 3 * 15 + 4 * 15 - 5 * 15

45 + 60 - 75

105 - 75

30

2. 3x + 4x - 5x

7x - 5x

2x

3. 3(x-2) + 4(x-2) - 5(x-2)

3x - 6 + 4x - 8 - 5x + 10

2x - 4

4. (I'm not typing that out)

7.5x + 19.5 + 10x + 26 - 12.5x - 32.5

5x+13

-hope it helps

8 0
1 year ago
1. ΔABC and ΔDEF are similar = Given
deff fn [24]
I think the answer is d. The slope of AC=Slope of DF.

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3 0
3 years ago
Read 2 more answers
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
I need help on my homework
soldi70 [24.7K]

Answer:

\displaystyle m\angle AED=32.5^\circ

Step-by-step explanation:

<u>Angles in a Circle</u>

An exterior angle of a circle is an angle whose vertex is outside a circle and the sides of the angle are secants or tangents of the circle.

Segments AE and DE are secants of the given circle. They form an exterior angle called AED.

The measure of an exterior angle is equal to half the difference of the measure of their intercepted arcs.

Intercepted arcs in the given circle are AD=113° and BC=48°. The exterior angle is:

\displaystyle m\angle AED=\frac{AD-BC}{2}

\displaystyle m\angle AED=\frac{113^\circ-48^\circ}{2}=\frac{65^\circ}{2}

\displaystyle m\angle AED=32.5^\circ

8 0
2 years ago
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