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erica [24]
4 years ago
12

Question 1 Pls answer Anyone who answers will be marked brainiest. Thanks u

Mathematics
1 answer:
goldfiish [28.3K]4 years ago
8 0

Answer:

2,4,5,and 7

Step-by-step explanation:

a term is either a single number or variable, or numbers and variables multiplied together.

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A sector of a circle has central angle θ=5π/6 and area 15π ft2. Find the radius of this circle.
Natali [406]
5pi/6 r^2 = 15pi

R^2 = 18

R= 3sqrt(3)
6 0
3 years ago
CITY |. Temperature (°F) St. Louis | 32 Montreal |. -18 Buffalo. | -8 Orlando. |. 68 Juneau. |. -26 By noon, the temperature In
Leno4ka [110]

Answer:

6°F

58°F

Step-by-step explanation:

From the data Given :

Temperature rise in Buffalo by noon time = 14°F

The initial temperature at Buffalo = - 8°F

Rise in temperature = 14°F

Therefore, temperature at noon = (-8 + 14)°F = 6°F

B.)

Temperature at 6am in St. Louis = 32°F

Temperature at 6am in Juneau = - 26°F

Temperature difference :

(32 - (-26))°F

32 + 26

= 58°F

Temperature in St. Louis is 58°F higher

8 0
3 years ago
Melissa’s father is 194 cm tall. He is 79 cm taller than Melissa. How tall is Melissa?
Kisachek [45]
Melissa is 115cm talk

4 0
3 years ago
Read 2 more answers
AMELIA IS SHARING 4 SLICES OF CHEESE WITH 5 FRIENDS. HOW MUCH CHEESE WILL EACH PERSON GET? EXPLAIN HOW YOU DECIDED.
Degger [83]

Answer:

well this is very easy all you have to do is divide 4 by 5 and you will get 4 with a remander of 1 or 0.8 or if this was a real life scenario then you could just take that extra slice and cut it in five ways. hope this helped! can i get brainiest plzz? im only ONE away from getting up to expert!

Step-by-step explanation:

3 0
3 years ago
Weights of the vegetables in a field are normally distributed. From a sample Carl Cornfield determines the mean weight of a box
pantera1 [17]
To find the z-score for a weight of 196 oz., use

z=\frac{x-\mu}{\sigma}=\frac{196-180}{8}=\frac{16}{8}=2

A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2.  Carl is wondering about the percentage of boxes with weights ABOVE z = 2.  The total area under the normal curve is 1, so subtract .9772 from 1.0000.

1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.

7 0
3 years ago
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