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3 years ago

15

There are 159 students who will be grouped into relay teams. Each team needs to have the same number of students. Should each te

am have 3, 5, 9, or 10 students? 3 students 10 students 5 students 9 students

1 answer:

Ber [7]3 years ago

8 0

3 students, 159 is divisible by 3. so it means that there will be 53 teams of 3

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What is the value of s?

Luba_88 [7]

Step-by-step explanation:

Sum of angles in a triangle = 180°.

=> s° + (s/7 + 3)° + (s/7 - 3)° = 180°

=> 9/7 * s° = 180°

=> s° = 180° * (7/9) = 140°

Hence the value of s is 140.

6 0

3 years ago

<br> PLS help me with this. And I am so sorry if im troubling you

dedylja [7]

Answer:

a=70 vertically opp angles

b=80 linear pair

c=100 corresponding angles

d= 80 linear pair with c

e= 180 -a-d

=180-70-80

=30

8 0

3 years ago

Yvette uses 6 games of tea leaves to make 24

Ludmilka [50]

The answer is 72 grams of tea leaves. 6 divided by 24 = 0.25 grams of tea leaves per 1 ounce of fluid tea. 0.25 x 288 = 72 grams of tea leaves.

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4 years ago

In this question, i is a unit vector due east and j is a unit vector due north. A cyclist rides at a speed of 4 m/s on a bearing

d1i1m1o1n [39]

<u>Answer:</u>

a) 1.04i + 3.86j

b) magnitude = 8; bearing = 302.7°

<u>Step-by-step explanation:</u>

a)

The first diagram represents the velocity vector of the cyclist.

To express this vector in the form xi + yj, we have to find the components of the vector in the horizontal (i) and vertical (j) directions.

If we consider the horizontal component of the vector to be $x$ , and the vertical component to be $y$ , then:

• horizontal component ⇒ $sin (15^{\circ}) = \frac{x}{4}$

⇒ $x = 4\space\ sin(15^{\circ})$

⇒ $x \approx \bf 1.04$

• vertical component ⇒ $cos(15^{\circ}) = \frac{y}{4}$

⇒ $y = 4 \space\ cos(15^{\circ})$

⇒ $y \approx \bf 3.86$

Now that we have the values of both the horizontal and vertical component, we can write the vector in the form of xi + yj:

vector ⇒ 1.04i + 3.86j

b)

The second diagram shows the first vector (red), the second vector (blue), and the resultant vector <em>v</em> (black). The dashed lines represent the components of the respective vectors.

To add two vectors given their magnitudes and direction, we have to add their components.

In order to find the horizontal and vertical components of the given vectors, we can use a method similar to that used above, so that:

○ For the first vector (magnitude 6):

• horizontal component ⇒ $x = 6 \space\ sin (60^{\circ})$

⇒ $\bf 5.2$

• vertical component ⇒ $y = 6 \space\ cos(60^{\circ})$

⇒ $y = \bf 3$

○ For the second vector (magnitude 2):

• horizontal component ⇒ $x = 2 \space\ cos (40^{\circ})$

⇒ $\bf 1.5$

• vertical component ⇒ $y = 2 \space\ sin(40^{\circ})$

⇒ $\bf 1.3$

Now we can add the respective components together:

v = 5.2i + 3j + 1.5i + 1.3j

⇒ (5.2 + 1.5)i + (3 + 1.3)j

⇒ 6.7i + 4.3j

∴ Magnitude of v ⇒ $|v| = \sqrt{(6.7)^2 + (4.3)^2}$

⇒ $|v| \approx \bf 8$

To find the bearing of <em>v</em>, we have to first calculate the angle marked $\alpha$ :

$tan \alpha = \frac{4.3}{6.7}$

⇒ $\alpha = tan^{-1}(\frac{4.3}{6.7})$

⇒ $\alpha = \bf 32.7^{\circ}$

∴ Bearing = 270° + 32.7°

= 302.7°

8 0

2 years ago

An article reports on an experiment in which each of five groups consisting of six rats was put on a diet with a different carbo

loris [4]

Answer:

H0 is rejected, Hence the true average DNA content is affected by the type of carbohydrate in the diet between samples

Step-by-step explanation:

<u> A B C D E</u>

<u>2.58 2.63 2.13 2.41 2.49</u>

∑A=2.58 ∑B=2.63 ∑C=2.13 ∑D=2.41 ∑E=2.49

<u>A² B² C² D² E²</u>

<u>6.6564 6.9169 4.5369 5.8081 6.2001</u>

∑A²=6.6564 ∑B²=6.9169 ∑C²=4.5369 ∑D²=5.8081 ∑E²=6.2001

Calculations

<u>Group A B C D E Total</u>

<u>N n1=1 n2=1 n3=1 n4=1 n5=1 n=5</u>

∑xi T1=2.58 T2=2.63 T3=2.13 T4=2.41 T5=2.49 ∑x=12.24

<u>∑x²i 6.6564 6.9169 4.5369 5.8081 6.2001 ∑x²=30.1184</u>

<u>Mean xi 2.58 2.63 2.13 2.41 2.49 Total2.448</u>

Σ Σ xij^2 = 185.4 given

Let k = the number of different samples = 5

n=n1+n2+n3+n4+n5=1+1+1+1+1=5

Overall ˉx=12.24/5 =2.448

∑x=T1+T2+T3+T4+T5=2.58+2.63+2.13+2.41+2.49=12.24----1

Correction Factor=(∑x)²/n=12.242/5=29.9635--------(2)

∑T²i/ni=(2.582/1+2.632/1+2.132/1+2.412/1+2.492/1)=30.1184------(3)

∑x²=∑x²1+∑x²2+∑x²3+∑x²4+∑x²5

=6.6564+6.9169+4.5369+5.8081+6.2001=30.1184-------(4)

ANOVA:

Step-1 : sum of squares between samples

SSB=(∑T2i/ni)-(∑x)2/n

=(3)-(2)

=30.1184-29.9635

=0.1549

Step-2 : Total sum of squares

SST=SSB-CF

= 185.4-2.488

=182.952

Step-3 : Sum of Squares Within Samples

SSW=∑x²-(∑T²i/ni)=SST- SSB=

= 182.952-0.1549=182.7971

Step-4 : variance between samples

MSB=SSB/k-1

=0.1549/4

=0.0387

Step-5 : variance within samples

MSW=SSW/n-k

=182.7971/5-5

=36.55942

Step-6 : test statistic F for one way ANOVA test

F=MSB/MSW

=0.0387/36.55942 = 0 .001058

<u>ANOVA table</u>

Source Sums Degrees Mean

of Variation of Squares of freedom Squares

SS DF MS F

B/w samples SSB = 0.1549 k-1 = 4 MSB = 0.0387 0.001

<u>Within samples SSW = 0 n-k = 0 MSW = 36.55942 </u>

<u>Total SST = 0.1549 n-1 = 4 </u>

H0 : The true average DNA content is not affected by the type of carbohydrate in the diet between samples

Ha : The true average DNA content is affected by the type of carbohydrate in the diet between samples

F(4,0) at 0.05 level of significance =0.5

As calculated F=0.001 >0.5

So, H0 is rejected, Hence the true average DNA content is affected by the type of carbohydrate in the diet between samples

6 0

4 years ago