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3 years ago

Help me with this please!!!!!!!!!

Mathematics

2 answers:

IrinaVladis [17]3 years ago

8 0

Answer:

d = 17.58

Step-by-step explanation:

d = √(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2

P1(0, 0, 0) P2(8, 7, 14)

Substitute the values in the above equation and you will get

d = √(8 - 0)^2 + (7 - 0)^2 + (14 - 0)^2

= √(64 + 49 + 196)

= √309

= 17.58

Anastasy [175]3 years ago

3 0

Answer:

The answer is 17.58

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Please help I will reward Brainly

Serggg [28]

The surface area increases by a factor of 4. 2^2= 4.
The area of the long side is 3×1= 3cm^2 on the small prism.
The area of the long side is 6×2= 12cm^2 on the large prism.
12/3= 4.

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3 years ago

Please help me explain as well. AD is 48. I will give brainliest.

otez555 [7]

Answer:

The area of the rectangle is 48

Step-by-step explanation:

We know that AE is 6 and ED is 8, since it forms a triangle, use the Pythagorean theorem to find AD, which is one side of the rectangle.

The Pythagorean theorem says that a^2 + b^2 = c^2.

a is AE, b is ED, and c is AD, so 6^2 + 8^2 = c^2, 36 + 64 = c^2, 100 = c^2, 10 = c, so AD is 10

The area of the triangle is half of AE times ED, so 1/2(6)(8) = 1/2(48) = 24, the area of the triangle is 24.

Since the area of a triangle is 1/2 of the base times the height, we can figure out the height:

A = 1/2bh, 24 = 1/2(10)h, 24 = 5h, 4.8 = h, so the height of the triangle (and rectangle) is 4.8.

The height of the triangle is also the height of the rectangle:

The area of the rectangle is 4.8*10 = 48

8 0

3 years ago

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr

Marianna [84]

Answer:

(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

(b) The value of t test statistics is 1.890.

(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) P-value of the test statistics is 0.0662.

Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men Women

The sample mean : 24.51 22.69

Sample standard deviation : 4.48 3.86

Sample size : 35 40

Let $\mu_1$ = mean number of times men order take-out dinners in a month.

$\mu_2$ = mean number of times women order take-out dinners in a month

(a) So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq$ 0 {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviation;

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 for men = 24.51

$\bar X_2$ = sample mean for women = 22.69

$s_1$ = sample standard deviation for men = 4.48

$s_2$ = sample standard deviation for women = 3.86

$n_1$ = sample of men = 35

$n_2$ = sample of women = 40

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2} }{35+40-2} }$ = 4.16

So, test statistics = $\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }$ ~ $t_7_3$

= 1.890

(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

P-value = P( $t_7_3$ > 1.89) = 0.0331

So, P-value for two tailed test is = 2 $\times$ 0.0331 = 0.0662