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AlekseyPX
2 years ago
8

Can someone help me with this math problem

Mathematics
2 answers:
swat322 years ago
4 0

Answer:

$14.00 per pound for A and $12.50 per pound for B and B is offering the better one since it has a lower cost per pound

Step-by-step explanation:

Shkiper50 [21]2 years ago
3 0

Answer:

Brand A is $14

Brand B is $12.5

Brand B is offering the better value. I know this because when you get the unit rate for Brand A it is $14. When you get the unit rate for Brand B you get $12.5. So You pay less money for Brand B then Brand A.

Step-by-step explanation:

For brand A you need to first divide 70 by 5 which equals to 14

Then you divide 105 by 12 which equals 12.5

The Last problem i gave to you in the answer box.

Hopefully you understood everything.

Have a good day

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john has $305, and he is spending $3 each day. which algebraic expression describes this situation, where d represents the numbe
Artist 52 [7]
The answer whould be like this

305 - 3d
8 0
3 years ago
Solve for X-4 1/6 = 9 1/3
Gala2k [10]

Answer:

  13 1/2

Step-by-step explanation:

X-4 1/6 = 9 1/3

Add 4 1/6 to each side

X-4 1/6+ 4 1/6 = 9 1/3 + 4 1/6

x = 9 1/3 + 4 1/6

  = 9 2/6 + 4 1/6

   13 3/6

    13 1/2

5 0
2 years ago
Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a
Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }

Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)

(Sigma/\sqrt{n} )= \frac{(X[bar]-Mu)}{Z}

Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}

Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}

n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2

n = (16*(\frac{1.2}{(270-264)}))^2

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

6 0
2 years ago
If Isabella runs 23 mile each day, how many miles will Isabella run in 6 days?
Kay [80]

Answer:

138 miles

Step-by-step explanation:

23x6=138

Pls give brainliest

6 0
2 years ago
Read 2 more answers
For what value of "a" will the product shown below have a purely imaginary value? (Show work please)
MakcuM [25]
If you expand out the brackets you get this,

(4+5i)(a+2i) = 4a + (5a)i + 8i - 10

The -10 comes from 5i * 2i.
Squaring i becomes -1.

Let's group the real stuff together,
and imaginary separately,

(4a - 10) + (5a + 8)i

For this to be purely imaginary,
the real part needs to be zero.

Therefore 4a - 10 = 0

Solve for a.
5 0
3 years ago
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