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3 years ago

Base your answer to the question on the diagrams shown below.

Mathematics

2 answers:

Galina-37 [17]3 years ago

5 0

The probability of drawing a red card and spinning a 1 are independent events, thus to get the probability of them occurring at the same time we shall have:
P(Red)×P(Spinning 1)
P(Red)=2/5
P(Spinning 1)=4/8=1/2
Thus:
P(Red)×P(Spinning 1)=1/2×2/5=1/5

zysi [14]3 years ago

3 0

Answer: $\frac{1}{5}$

Step-by-step explanation:

From the given picture, the total number of cards = 5

The number of red card = 2

Thus, the probability of drawing a red card P(red)= $\frac{2}{5}$

Also, total number of digits on spinner = 8

Number of 1's=4

Thus, the probability of spinning a 1 P( spinning 1)= $\frac{4}{8}=\frac{1}{2}$

Since both the events of drawing a red card and spinning a 1 are independent events, therefore, the probability of drawing a red card and then spinning a 1= $\text{ P(red)}\times\text{P( spinning 1)}=\frac{2}{5}\times\frac{1}{2}=\frac{1}{5}$

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The area of a triangle whose sides are 3cm, 4cm and 5cm respectively.

irakobra [83]

<h2>3x4x5=60cms </h2><h2>60cm is the correct answer to the question </h2>

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3 years ago

Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute

valkas [14]

Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

= 3 - 1 = 2

d.f.D = N - k

= 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust Jet red Cloudtran

$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

$s_1^2=19.96154$ $s_2^2=14.68681$ $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

$=\frac{51+51+...+49+49}{35}$

= 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

$=\frac{127.1143}{2}$

= 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

3 0

3 years ago

Teresa bought 3boxes of chocolate each box has 9 pieces of chocolate inside teresa gave away 15 pieces of chocolate how many pie

lora16 [44]

Hey There!

3×9=27
27-15=12
Teresa has 12 pieces of Chocolate left

Hope This Helps!!!!

4 0

3 years ago

Read 2 more answers

Trapezoids and kites

lisov135 [29]

QUESTION 3

The sum of the interior angles of a kite is $360\degree$ .

$\Rightarrow 36\degree +70\degree+m$ .

$\Rightarrow 106\degree+m$ .

$\Rightarrow m$ .

But the two remaining opposite angles of the kite are congruent.

$\Rightarrow m$

$\Rightarrow m$ .

$\Rightarrow 2m$ .

$\Rightarrow m$ .

QUESTION 4

RH is the hypotenuse of the right triangle formed by the triangle with side lengths, RH,12, and 20.

Using the Pythagoras Theorem, we obtain;

$|RH|^2=12^2+20^2$

$|RH|^2=144+400$

$|RH|^2=544$

$|RH|=\sqrt{544}$

$|RH|=4\sqrt{34}$

QUESTION 5

The given figure is an isosceles trapezium.

The base angles of an isosceles trapezium are equal.

Therefore $m$

QUESTION 6

The measure of angle Y and Z are supplementary angles.

The two angles form a pair of co-interior angles of the trapezium.

This implies that;

$m$

$\Rightarrow m$

QUESTION 7

The sum of the interior angles of a kite is $360\degree$ .

$\Rightarrow 48\degree +110\degree+m$ .

$\Rightarrow 158\degree+m$ .

$\Rightarrow m$ .

But the two remaining opposite angles are congruent.

$\Rightarrow m$

$\Rightarrow m$ .

$\Rightarrow 2m$ .

$\Rightarrow m$ .

QUESTION 8

The diagonals of the kite meet at right angles.

The length of BC can also be found using Pythagoras Theorem;

$|BC|^2=4^2+7^2$

$\Rightarrow |BC|^2=16+49$

$\Rightarrow |BC|^2=65$

$\Rightarrow |BC|=\sqrt{65}$

QUESTION 9.

The sum of the interior angles of a trapezium is $360\degree$ .

$\Rightarrow m$ .

But the measure of angle M and K are congruent.

$\Rightarrow m$ .

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murzikaleks [220]

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3 years ago