Can someone please answer these two questions

Please help! :) (Whoever answers first I'll give Brainlest!)

Irina-Kira [14]

Answer:

87

Step-by-step explanation:

Do what's inside the paranethesis first

2+(9*3)=2+27=29

29*3=87

7 0

3 years ago

PLEASE SHOW FULL SOLUTIONS!!! WILL MARK BRAINLIEST!!! SHOW ALL STEPS!!! THANK YOU!!!

ELEN [110]

9514 1404 393

Answer:

a) BE = 5; DE = 6; EF = 4

b) ∠EFC ≅ ∠DA.F ≅ ∠BDE or <em>b = e = f = i</em>

Step-by-step explanation:

Each short segment is the same length as the marked one it is parallel to.

E is the midpoint of BC, so BE = EC = 5.

ADEF is a parallelogram, so DE = A.F = 6.

D is the midpoint of AB, so AD = DB = 4. ADEF is a parallelogram, so ...

EF = AD = 4

__

As we have noted, AB║EF and DE║A.F, so corresponding angles and alternate interior angles are congruent. <em>b = e = f = i</em>

5 0

3 years ago

What is the value of a in the equation −14a−5=−12?

-BARSIC- [3]

-14a-5=-12
+5 +5
-14a=-7
Divide -7 by -14a
a=2

7 0

3 years ago

James has 14{,}500\text { g}14,500 g14, comma, 500, start a text, space, g, end text of sand in his sandbox. He brings home anot

mariarad [96]

Question:

James has 14,500g of sand in his sandbox. He brings home another 7,400g of sand from the beach to add to his sandbox. How many kilograms of sand does James have in his sandbox now?

Answer:

The quantity of sandbox James has is 21.9kg

Step-by-step explanation:

Given

Initial quantity of sandbox = 14,500 g

Additional quantity = 7,400 g

Required

Total quantity of sandbox in kg

Tp get the total quantity of sandbox, we simply add in the initial and additional quantity of sandbox together;

This is represented mathematically as follows;

Total = Initial Quantity + Additional Quantity

Total = 14,500 g + 7,400 g

Total = 21,900 g

But the question says the answer should be in kg

1000 g is equivalent to 1 kg

21,900 g will be equivalent to $\frac{21,900}{1,000} kg$

$21,900g = \frac{21,900}{1,000} kg$

$21,900g = 21.9 kg$

Hence, the quantity of sandbox James has is 21.9kg

7 0

3 years ago

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in, and a standard deviation given by

kirza4 [7]

Answer: a) The probability is approximately = 0.5793

b) The probability is approximately=0.8810

Step-by-step explanation:

Given : Mean : $\mu= 62.5\text{ in}$

Standard deviation : $\sigma = \text{2.5 in}$

a) The formula for z -score :

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

Sample size = 1

For x= 63 in. ,

$z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{1}}}=0.2$

The p-value = $P(z$

$0.5792597\approx0.5793$

Thus, the probability is approximately = 0.5793

b) Sample size = 35

For x= 63 ,

$z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{35}}}\approx1.18$

The p-value = $P(z$

$= 0.8809999\approx0.8810$

Thus , the probability is approximately=0.8810.

6 0

4 years ago