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sattari [20]
2 years ago
12

0.5(400 + d) =4.25 d=​

Mathematics
2 answers:
agasfer [191]2 years ago
7 0

Answer:

pleas help me asap please!!!!!!

pleas help me asap please!!!!!!

Step-by-step explanation:

pleas help me asap please!!!!!!

qaws [65]2 years ago
6 0

Answer:

d = 53.33333

Step-by-step explanation:

Solve for d on calculator

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In a litter of 7 kittens, each kitten weighs less than 3.5 ounces. Find all the possible values of the combined weights of the k
Dima020 [189]

Let's use w to symbolize the weight of kittens. Given by the statement "each kitten weighs less than 3.5 ounces" we know that

1*w < 3.5

We can multiply both sides of the inequality by 7 to determine the total weights

7*(1*w) < 7*(3.5)

7*w < 24.5

Since there are 7 kittens, the combined weight of the kittens is 7w, therefore the above expression could be read as "The combined weight of the kittens is less than 24.5 ounces"

6 0
3 years ago
3. What is the slope of the line through the points (2, 5) and (6, 13)?
White raven [17]

Answer:

m = 2

Step-by-step explanation:

m=\frac{rise}{run}=\frac{13-5}{6-2}=\frac{8}{4}=\boxed{2}

Hope this helps.

8 0
3 years ago
If miley has 200 pencils and Katie has 300 more pencils then how many more does Katie have then miley
Yakvenalex [24]

Answer: 300

Step-by-step explanation: 300+200=500

then 500-200=300

3 0
3 years ago
Read 2 more answers
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

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3 years ago
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What is your question I will answer it if I can
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