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kupik [55]
2 years ago
11

Ill give brainllest what is 18+26+37+36+8+28?

Mathematics
2 answers:
Mekhanik [1.2K]2 years ago
5 0

Answer:

18+26+37+36+8+28 = 153

hope this helps!

Rashid [163]2 years ago
4 0

Answer:

153

Step-by-step explanation:

You might be interested in
7 1/3 ∙ x = 1.6 ÷ 6/11<br><br> x = __ <br><br> Please Help!
ANTONII [103]

Answer:

x= 0.4

Step-by-step explanation:

We need to solve and find the value of x

7 1/3x = 1.6 ÷ 6/11

Converting mixed fraction into improper fraction

22/3 x = 1.6 ÷ 6/11

Converting ÷ sign by X and reciprocating the term

22/3 x = 1.6 * 11/6

Now divide both sides by 3/22

x = 1.6 * 11/6 * 3/22

x = 1.6 * 1/2 * 1/2

x = 1.6/4

x = 0.4

S0, x= 0.4

7 0
3 years ago
Put these fractions and decimals in least to greatest<br><br> 8 1/9,8.117,8.28,163/20
suter [353]
Turn them all into decimals so you can clearly see the difference between them

8 1/9 = 8.11
8.117
8.28
163/20 = 8.15

8 1/9, 8.117, 163/20, 8.28 is your answer
8 0
2 years ago
Help me please look at the file
Alchen [17]

Answer:

350 will be your answer :)

Step-by-step explanation:

so i did 70 divided by 2 because 140 divided by 70 = 2

and 70 divded by 2 IS 35!

But wait that isn't what it's asking for right?

then i did 35 TIMES 10 cuz that's what it;s asking for

i got 350 and so should you so in conclusion 350 is your answer

Hopefully this helped!

8 0
2 years ago
Read 2 more answers
Jake is making cookies. To make 30 cookies, his recipe requires 4 eggs. Using this recipe, what is the total number of eggs he w
TiliK225 [7]

Answer:

16 eggs

Step-by-step explanation:

<u>Steps to answering this question </u>

  1. determine how many eggs would be required to make one cookies
  2. Multiply the answer gotten above by the number of cookies

Number of eggs needed for one cookie = 4/30 = 0.1333

Number of eggs needed for 120 cookies = 0.1333 x 120 = 16 eggs

8 0
2 years ago
4) Dan Ariely and colleagues have conducted some extremely interesting studies on cheating (see, for example Mazar, Amir, &amp;
lianna [129]

I've attached the complete question.

Answer:

Only participant 1 is not cheating while the rest are cheating.

Because only participant 1 has a z-score that falls within the 95% confidence interval.

Step-by-step explanation:

We are given;

Mean; μ = 3.3

Standard deviation; s = 1

Participant 1: X = 4

Participant 2: X = 6

Participant 3: X = 7

Participant 4: X = 0

Z - score for participant 1:

z = (x - μ)/s

z = (4 - 3.3)/1

z = 0.7

Z-score for participant 2;

z = (6 - 3.3)/1

z = 2.7

Z-score for participant 3;

z = (7 - 3.3)/1

z = 3.7

Z-score for participant 4;

z = (0 - 3.3)/1

z = -3.3

Now from tables, the z-score value for confidence interval of 95% is between -1.96 and 1.96

Now, from all the participants z-score only participant 1 has a z-score that falls within the 95% confidence interval.

Thus, only participant 1 is not cheating while the rest are cheating.

8 0
3 years ago
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