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LuckyWell [14K]
3 years ago
7

Charmaine is on the swim team. Each day she swims 950m . How far does she swim in 5 days?

Mathematics
2 answers:
Roman55 [17]3 years ago
3 0
950m * 5 days = 4750 m = 4.75 km
Trava [24]3 years ago
3 0
4750m
..............................
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In a right triangle, the length of one leg is 6 units. The length of the other leg is 8 units. What is the length of the hypoten
lozanna [386]
hypotenuse =  \sqrt{6^2+8^2}= \sqrt{36+64}= \sqrt{100}=10 \ units
7 0
3 years ago
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Consider the given right triangle. If angle B = 43 degrees and c = 26.7 meters, find a.
Nady [450]

Answer: Side a equals 19.5 metres

Step-by-step explanation: Consider the right angled triangle as shown in the picture attached. The triangle has been drawn with angle measuring 43 degrees, side c (line AB) measuring 26.7 m and side a (line CB) is yet unknown.

A right angled triangle can be solved if at least one side and an angle are available. In this question we shall apply the trigonometric ratios since we have one angle which shall be the reference angle (43°). Also we have an hypotenuse (the side facing the right angle) and an unknown side which is the adjacent (which lies between the right angle and the reference angle).

Cos B = Adjacent/Hypotenuse

Cos 43 = a/26.7

Cos 43 x 26.7 = a

0.7314 x 26.7 = a

19.52714 = a

a ≈ 19.5  (rounded to the nearest tenth)

Therefore the length of side a equals 19.5 metres.

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3 years ago
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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

<em>Let </em>\mu_1<em> = mean number of times men order take-out dinners in a month.</em>

<em />\mu_2<em> = mean number of times women order take-out dinners in a month</em>

(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 <u>Two-sample t test statistics</u> 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, <u>test statistics</u>  =  \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 <u>we fail to reject our null hypothesis</u>.

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 = <u>0.0662</u>

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Answer:

0.15

Step-by-step explanation:

3divided by 2

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3 years ago
Name a three-dimensional shape that has four triangular faces and one rectangular face. Name a three-dimensional shaoe that has
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The answers would be pyramid and triangular prism.

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