Answer:
C. -10Step-by-step explanation:
[tex]hope \: it \: helps[/tex]
CarryOnLearning
Instructions: State what additional information is required in order
to know that the triangles in the image below are congruent for the
reason given
Reasory. SAS Postulate
Answer:
HJ = FG
Step-by-step explanation:
SAS means side - (included) angle - side.
we have one angle confirmed (at H and at G).
we have actually one side confirmed (HG), because the graphic shows that this side is shared between the triangles. so, implicitly it is not only congruent but really identical.
so, we need the confirmation of the second side enclosing the confirmed angle.
Leanne is planning a bridal shower for her best friend. At the party, she wants to serve 33 beverages, 33 appetizers, and 22 desserts, but she does not have time to cook. She can choose from 1313 bottled drinks, 77 frozen appetizers, and 1313 prepared desserts at the supermarket. How many different ways can Leanne pick the food and drinks to serve at the bridal shower
Answer:
She can pick the food and drinks in 780,780 different ways.
Step-by-step explanation:
The drinks, appetizers and desserts are independent of each other, so the fundamental counting principle is used.
Also, the order in which the beverages, the appetizers and the desserts are chosen is not important, which means that the combinations formula is used to solve this question.
Fundamental counting principle:
States that if there are p ways to do a thing, and q ways to do another thing, and these two things are independent, there are p*q ways to do both things.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
Beverages:
3 from a set of 13. So
[tex]C_{13,3} = \frac{13!}{3!10!} = 286[/tex]
Appetizers:
3 from a set of 7, so:
[tex]C_{7,3} = \frac{7!}{3!4!} = 35[/tex]
Desserts:
2 from a set of 13, so:
[tex]C_{13,2} = \frac{13!}{2!11!} = 78[/tex]
How many different ways can Leanne pick the food and drinks to serve at the bridal shower?
286*35*78 = 780,780
She can pick the food and drinks in 780,780 different ways.
Suppose g(x) = f( x +2) - 3. Which statement best compares the graph of g(x) with the graph of f(x)? A. The graph of g(x) is shifted 2 units left and 3 units up. B. The graph of g(x) is shifted 2 units right and 3 units down. C. The graph of g(x) is shifted 2 units left and 3 units down. D. The graph of g(x) is shifted 2 units right and 3 units up.
Given:
The function is:
[tex]g(x)=f(x+2)-3[/tex]
To find:
The statement that best compares the graph of g(x) with the graph of f(x).
Solution:
The transformation is defined as
[tex]g(x)=f(x+a)+b[/tex] .... (i)
Where, a is horizontal shift and b is vertical shift.
If a>0, then the graph shifts a units left and if a<0, then the graph shifts a units right.
If b>0, then the graph shifts b units up and if b<0, then the graph shifts b units down.
We have,
[tex]g(x)=f(x+2)-3[/tex] ...(ii)
On comparing (i) and (ii), we get
[tex]a=2[/tex]
[tex]b=-3[/tex]
Therefore, the graph of g(x) is shifted 2 units left and 3 units down.
Hence, the correct option is C.
There are two types of fish in a lake. These are carp and pike.
In a netted area of the lake 120 carp and 16 pike were caught
In the whole lake it is estimated there are 34 000 fish.How many pike are there?
Answer:pike are 250 ; from 120+16=136 and 34000/136 is 250
Step-by-step explanation:
What is the value of x that makes l1||l2
A. 35
B. 25
C. 37
D. 18
Answer:
B
Step-by-step explanation:
For l1 and l2 to be parallel, these two angles need to be equal. 3x-15=2x+10, x=25
Many freeways have service (or logo) signs that give information on attractions, camping, lodging, food, and gas services prior to off-ramps. These signs typically do not provide information on distances. An article reported that in one investigation, six sites along interstate highways where service signs are posted were selected. For each site, crash data was obtained for a three-year period before distance information was added to the service signs and for a one-year period afterward. The number of crashes per year before and after the sign changes were as follows.
Before 13 22 65 123 56 63
After 14 21 43 84 75 72
1. The article included the statement "A paired t-test was performed to determine whether there was any change in the mean number of crashes before and after the addition of distance information on the signs." Carry out such a test. (Note: The relevant normal probability plot shows a substantial linear pattern.)
a. State and test the appropriate hypotheses. (Use α = 0.05.)
b. Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)
t = _____
p-value = _____
c. State the conclusion in the problem context.
A. Fail to reject H0. The data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.
B. Reject H0. The data suggests a significant mean difference in the average number of accidents after information was added to road signs.
C. Fail to reject H0. The data suggests a significant mean difference in the average number of accidents after information was added to road signs.
D. Reject H0. The data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.
2. If a seventh site were to be randomly selected among locations bearing service signs, between what values would you predict the difference in the number of crashes to lie? (Use a 95% prediction interval. Round your answers to two decimal places.)
Answer:
Test statistic = 0.63
Pvalue = 0.555
A. Fail to reject H0. The data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.
Step-by-step explanation:
Given :
Before 13 22 65 123 56 63
After_ 14 21 43 84 75 72
To perform a paired t test :
H0 : μd = 0
H1 : μd ≠ 0
We obtain the difference between the two dependent sample readings ;
Difference, d = -1, 1, 22, 39, -19, -9
The mean of difference, Xd = Σd/ n = 33/6 = 5.5
The standard deviation, Sd = 21.296 (calculator).
The test statistic :
T = Xd ÷ (Sd/√n) ; where n = 6
T = 5.5 ÷ (21.296/√6)
T = 5.5 ÷ 8.6940555
T = 0.6326
The Pvalue : Using a Pvalue calculator ;
df = n - 1 = 6 - 1 = 5
Pvalue(0.6326, 5) = 0.5548
Decision region :
Reject H0 ; If Pvalue < α; α = 0.05
Since 0.5548 > 0.05 ; we fail to reject the Null and conclude that the data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.
giving brainliest! :)
Answer:
239=5
478=10
956=20
Step-by-step explanation:
Gianna's car can travel 478 mi with 10 gallons of gas
so, 47.8 mi with 1 gallon
by dividing 239 by 47.8 miles/gallon we get the answer 5 miles.
if we multiply 20 gallons by 47.8 mi/gallon #e get the answer 956 miles.
easy weasy
A worker in the automobile industry works an average of 43.7 hours per week. Assume the distribution is normal with a standard deviation of 1.6 hours.
(i) What is the probability that a randomly selected automobile worker works less than 40 hours per week?
(ii) If 15 automobile workers are randomly selected, what is the probability that the sample mean of working time is more than 45 hours per week?
Answer:
The solution is:
(1) 0.0104
(2) 0.0008
Step-by-step explanation:
Given:
Mean,
[tex]\mu = 43.7[/tex]
Standard deviation,
[tex]\sigma = 1.6[/tex]
(1)
⇒ [tex]P(X<40) = P(\frac{x-\mu}{\sigma}<\frac{40-43.7}{1.6} )[/tex]
[tex]=P(z< - 2.3125)[/tex]
[tex]=P(z<-2.31)[/tex]
[tex]=0.0104[/tex]
(2)
As we know,
n = 15
⇒ [tex]P(\bar X > 45)= P(\frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n} } } >\frac{45-43.7}{\frac{1.6}{\sqrt{15} } } )[/tex]
[tex]=P(z> 3.15)[/tex]
[tex]=1-P(z<3.15)[/tex]
[tex]=1-0.9992[/tex]
[tex]=0.0008[/tex]
Twice the difference of a number and 9 equals 5.
Answer:
7
Step-by-step explanation:
if the number is x then you know 2x-9=5
2x=14
x=7
so the number is 7
Please help quicklyyy!!!
Answer:
Its the 3 one
Step-by-step explanation:
The physical plant at the main campus of a large state university recieves daily requests to replace fluorescent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 45 and a standard deviation of 3. Using the empirical rule, what is the approximate percentage of lightbulb replacement requests numbering between 42 and 45?
Do not enter the percent symbol.
ans = %
Answer:
34%
Step-by-step explanation:
Given that the distribution of daily light bulb request replacement is approximately bell shaped with ;
Mean , μ = 45 ; standard deviation, σ = 3
Using the empirical formula where ;
68% of the distribution is within 1 standard deviation from the mean ;
95% of the distribution is within 2 standard deviation from the mean
Lightbulb replacement numbering between ;
42 and 45
Number of standard deviations from the mean /
Z = (x - μ) / σ
(x - μ) / σ < Z < (x - μ) / σ
(42 - 45) / 3 = -1
This lies between - 1 standard deviation a d the mean :
Hence, the approximate percentage is : 68% / 2 = 34%
Solve the equation for x 11x=110
15. The height and yolume of a cylinder are 4cm and 616cm respectively. Calculate the diameter of the base. (take t = 27 쪽 A. 7cm B. 154cm C. 14cm D. 64cm
Step-by-step explanation:
Given that,
Height of cylinder = 4 cmVolume of cylinder = 616 cm³To find,
Diameter of the base = ?Firstly we'll find the base radius of the cylinder.
[tex]\longmapsto\rm{V_{(Cylinder)} = \pi r^2h}\\[/tex]
According to the question,
[tex]\longmapsto\rm{616= \dfrac{22}{7} \times r^2 \times 4}\\[/tex]
[tex]\longmapsto\rm{616 \times 7 = 22 \times r^2 \times 4}\\[/tex]
[tex]\longmapsto\rm{4312 = 88 \times r^2 }\\[/tex]
[tex]\longmapsto\rm{\cancel{\dfrac{4312}{88}} = r^2 }\\[/tex]
[tex]\longmapsto\rm{49 = r^2 }\\[/tex]
[tex]\longmapsto\rm{\sqrt{49} = r }\\[/tex]
[tex]\longmapsto\rm{7 \; cm = r }\\[/tex]
Now,
[tex]\longmapsto\rm{Diameter = 2r }\\[/tex]
[tex]\longmapsto\rm{Diameter = 2(7 \; cm) }\\[/tex]
[tex]\longmapsto\bf{Diameter = 14 \; cm}\\[/tex]
The required answer is 14 cm.
14 over 17 as a decimal rounded to the nearest tenth
Step-by-step explanation:
14/17 is 0.82352941176
To the nearest tenth is 0.8
What is the length of the arc of a circle of diameter 8 meters subtended by a central angle of
3pi/4 radians?
Answer:
9.42 meters
Step-by-step explanation:
diameter = 8 m
radius = 4 m
Length of arc = radius * central angle in radians
= 4 * 3[tex]\pi[/tex] / 4
= 12[tex]\pi[/tex] / 4
= 3[tex]\pi[/tex]
= 66 / 7
= 9.42 m
The length of the arc of a circle of diameter 8 meters subtended by a central angle of 3pi/4 radians is 9.42 meters.
What is the arc of the circle?The arc period of a circle can be calculated with the radius and relevant perspective using the arc period method.
expalination:
⇒angle= arc/radius
= 3pi/4 radians
diameter = 8 m
radius = 4 m
Length of arc = radius * central angle in radians
⇒4 * 3 / 4
⇒12 / 4 = 3
⇒66 / 7
⇒9.42 m
Hence the arc of a circle is 9.42 m.
Learn more about arc of a circle here:-https://brainly.com/question/2005046
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A ladder leans against the side of the a house. The ladder is 19 feet long and forms an angle of elevation of 75 degree when leaned against the house. How far away from the house is the ladder? Round your answer to the nearest tenth.
Answer: 18.35259
Hope it helps!
find the supplement of 158 degrees and 17 minutes
Answer:
supplement of 158 degree
x+158=180
x=180-158
x=22 degree.
Step-by-step explanation:
A rectangular prism has volume 1,088 ft3 and height 8 ft. What is the area of the base of the prism?
a. 146 ft2
c. 136 ft2
b. 1,080 ft2
d. 1,096 ft2
We know
[tex]\boxed{\sf Volume=Area\:of\:Base\times Height}[/tex]
[tex]\\ \sf\longmapsto Area\:of\:Base=\dfrac{Volume}{Height}[/tex]
[tex]\\ \sf\longmapsto Area\:of\:Base=\dfrac{1088}{8}[/tex]
[tex]\\ \sf\longmapsto Area\;of\:base=136ft^2[/tex]
PLEASE help me solve this!!!!
1. log2^64
2. log11^121
3. log8^512
4. log4^1/16
Here are all of the answers to these four questions:
1. log2^64=6, or 2 to the power of 6=64.
2. log11^121=2, or 11²=121.
3. log8^512=3, or 8³=512.
4. log4^1/16=-2, or 4 to the negative second (-2) power.
I hope that this answered your question.
Yeah, so the answer is 6, 2, 3, and -2.
A rhombus has an area of 5 square meters and a side length of 3 meters. In another similar rhombus, the length of a side is 9 meters. What is the area of the second rhombus?
(A) 30 square meters
(B) 45 square meters
(C) 60 square meters
(D) 75 square meters
Hence the area of the second rhombus is 45 square meters
The area of a rhombus is expressed as
A = base * height
For the rhombus with an area of 5 square meters and a side length of 3 meters
Height = Area/length
Height = 5/3 metres
Since the length of a similar rhombus is 9meters, the scale factor will be expressed as;
k = ratio of the lengths = 9/3
k = 3
Height of the second rhombus = 3 * height of the first rhombus
Height of the second rhombus = 3 * 5/3
Height of the second rhombus = 5 meters
Area of the second rhombus = length * height
Area of the second rhombus = 5 * 9
Area of the second rhombus = 45 square meters
Hence the area of the second rhombus is 45 square meters
Learn more here: brainly.com/question/20247331
The correct option is option B;
(B) 45 square meters
The known parameters in the question are;
The area of the rhombus, A₁ = 5 m²
The length of one of the sides of the rhombus, a = 3 m
The length of a side in a similar rhombus, b = 9 m
The unknown parameter;
The area of the second rhombus
Strategy or method;
We have that two shapes are similar if their corresponding sides are proportional
From the above statement we get that the ratio of the areas of the two shapes is equal to the square of the ratio of the lengths of the corresponding sides of the two shapes of follows;
[tex]\begin{array}{ccc}Length \ Ratio&&Area \ Ratio\\\dfrac{a}{b} &&\left (\dfrac{a}{b} \right)^2 \\&&\end{array}[/tex]
Let the area of the second rhombus be A₂, we get;
[tex]Area \ ratio = \dfrac{A_1}{A_2} = \left( \dfrac{a}{b} \right)^2[/tex]
Where;
a = 3 m, b = 9 m, and A₁ = 5 m², we get;
[tex]Area \ ratio = \dfrac{5 \ m^2}{A_2} = \left( \dfrac{3 \, m}{9 \, m} \right)^2 = \dfrac{1}{9}[/tex]
Therefore;
9 × 5 m² = A₂ × 1
A₂ = 45 m²
The area of the second rhombus, A₂ = 5 m².
Learn more about scale factors here;
https://brainly.com/question/20247331
Write the equation of the line in fully simplified slope-intercept form.
Answer:
y = 6/5x-1
Step-by-step explanation:
We have two points so we can find the slope
(-5,-7) and (5,5)
The slope is
m = ( y2-y1)/(x2-x1)
= ( 5- -7)/( 5 - -5)
= (5+7)/(5+5)
= 12/10
= 6/5
The slope intercept form of a line is
y = mx+b
y = 6/5x+b
Using the point (5,5)
5 = 6/5(5)+b
5=6+b
b=-1
y = 6/5x-1
There is 60% chance of making $12,000, 10% chance of breaking even and 30% chance of losing $6,200. What is the expected value of the purchase?
Answer:
$5,340
Step-by-step explanation:
Given :
Making a probability distribution :
X : ___12000 ____0 _____-6200
P(X) __ 0.6 _____ 0.1 _____ 0.3
Tge expected value of the purchase si equal to the expected value or average, E(X) :
E(X) = ΣX*p(X)
E(X) = (12000 * 0.6) + (0 * 0.1) + (-6200 * 0.3)
E(X) = 7200 + 0 - 1860
E(X) = $5,340
Using the identity. (a - b) ²= (a² - 2ab + b²), evaluate 699²
Step-by-step explanation:
hope it helps you.......
[tex]\\ \sf\longmapsto 699^2[/tex]
[tex]\\ \sf\longmapsto (700-1)^2[/tex]
[tex]\\ \sf\longmapsto 700^2-2(700(1)+(1)^2[/tex]
[tex]\\ \sf\longmapsto 490000-1400+1[/tex]
[tex]\\ \sf\longmapsto 488600+1[/tex]
[tex]\\ \sf\longmapsto 488601[/tex]
An adult can lose or gain two pounds of water ina course of a day. Assume that the changes in water weight isuniformly distributed between minus two and plus two pounds in aday. What is the standard deviation of your weight over a day?
Answer:
The standard deviation of your weight over a day is of 1.1547 pounds.
Step-by-step explanation:
Uniform probability distribution:
An uniform distribution has two bounds, a and b, and the standard deviation is:
[tex]S = \sqrt{\frac{(b-a)^2}{12}}[/tex]
Assume that the changes in water weight is uniformly distributed between minus two and plus two pounds in a day.
This means that [tex]a = -2, b = 2[/tex]
What is the standard deviation of your weight over a day?
[tex]S = \sqrt{\frac{(2 - (-2))^2}{12}} = \sqrt{\frac{4^2}{12}} = \sqrt{\frac{16}{12}} = 1.1547[/tex]
The standard deviation of your weight over a day is of 1.1547 pounds.
Five trucks are to be transported on a ship. Each one weighs 3200 kg and comes
with 8 tyres which weigh 125 kg each. what is the total weight
Total No of trucks: 5
Weight of trucks: 3200Kg
Total weight of trucks: 3200×5
= 16000kg
Total no of tyres = 5 ×8
= 40
Weight of each tyre = 125kg
Total weight of tyres = 125 × 40
= 5000Kg
The total weight of trucks and tyres: 16000 + 5000
= 21000Kg
Answered by Gauthmath must click thanks and mark brainliest
The central angle in a circle of radius 6 meters has an intercepted arc length of 10 meters. Find the measure of the angle in radians and in degrees
Answer:
The central angle is 5/3 radians or approximately 95.4930°.
Step-by-step explanation:
Recall that arc-length is given by the formula:
[tex]\displaystyle s = r\theta[/tex]
Where s is the arc-length, r is the radius of the circle, and θ is the measure of the central angle, in radians.
Since the intercepted arc-length is 10 meters and the radius is 6 meters:
[tex]\displaystyle (10) = (6)\theta[/tex]
Solve for θ:
[tex]\displaystyle \theta = \frac{5}{3}\text{ rad}[/tex]
The central angle measures 5/3 radians.
Recall that to convert from radians to degrees, we can multiply by 180°/π. Hence:
[tex]\displaystyle \frac{5\text{ rad}}{3} \cdot \frac{180^\circ}{\pi \text{ rad}} = \frac{300}{\pi}^\circ\approx 95.4930^\circ[/tex]
So, the central angle is approximately 95.4930°
find the missing side round to the nearest tenth
Answer:
[tex]cos49 = \frac{26}{x} \\ x = 39.6305[/tex]
What is the measure of x?
Answer:
22
Step-by-step explanation:
This is a right angle so the sum of those would be equal to 90 degrees
x + 7 + 3x - 5 = 90 add like terms
4x + 2 = 90 subtract 2 from both sides
4x = 88 divide both sides by 4
x = 22
The graph shows the distribution of lengths of songs (in seconds). The distribution is approximately Normal, with a mean of 227 seconds and a standard deviation of 31 seconds.
A graph titled Song length has length (seconds) on the x-axis, going from 103 to 351 in increments of 31. The highest point of the curve is at 227.
What percentage of songs have lengths that are within 31 seconds of the mean?
34%
68%
95%
99.7%
its everything between 196 and 258 seconds (at max 31secs away from the mean). imagine straight upward lines separating this area from the rest.
34% would be way too low, 95 and above way too much.
only 68% is remotely plausible.
Find the first six terms of the sequence.
a1=- 6, an= 4 • an-1
Answer:
I think it's 3
Step-by-step explanation:
because an=4 and the question is
an-1
=4-1
=3