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3 years ago

Solve the proportion for the variable. W/5=49/35

Mathematics

1 answer:

hammer [34]3 years ago

7 0

Answer:

Step-by-step explanation:

So if w/5 = 49/35, then you can take w/5 and make the denominator 35 by multiplying the top and bottom by 7:

w/5 * 7/7 = 7w/35

So then you have:

7w/35 = 49/35

Which means:

7w = 49

w = 7

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3 years ago

Weight gain during pregnancy. In 2004, the state of North Carolina released to the public a large data set containing informatio

pychu [463]

Answer:

1. B. H0: μ1−μ2=0, HA: μ1−μ2≠0

2. z=1.2114

3. P-value=0.2257

4. Do not reject H0

Step-by-step explanation:

We have to perfomr an hypothesis test to see if there is strong evidence that there is a significant difference between the two population means.

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2\neq0$

Being μ1 the mean average gain for younger mothers and μ2 the mean average gain for mature mothers.

(NOTE: we are comparing means, not proportions, as it is the random variable is the weight gain).

As we are claiming "strong evidence", the level of significance will be 0.01.

For younger mothers, the sample size is n1=840, the sample mean is 30.7 and the sample standard deviation is s1=14.91.

For mature mothers, the sample size is n2=132, the sample mean is 29.15 and the sample standard deviation is s2=13.46.

The difference between means is

$M_d=\mu_1-\mu_2=30.7-29.15=1.55$

The standard error of the difference between means is

$s_M=\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}=\sqrt{\dfrac{14.91^2}{840}+\dfrac{13.46^2}{132}}=\sqrt{ 0.2647+1.3725}=\sqrt{1.6372}\\\\\\s_M=1.2795$

Then, the statistic can be calculated as:

$z=\dfrac{M_d-(\mu_1-\mu_2)}{s_M}=\dfrac{1.55-0}{1.2795}=1.2114$

The P-value for this z-statistic in a two tailed test is:

$P-value=2P(z>1.2114)=0.2257$

As the P-value is greater than the significance level, the null hypothesis failed to be rejected.

There is no enough evidence to claim that the real average weight gain differs from mature and youger mothers.

5 0

4 years ago

I just need an equation for this

34kurt

Um, I think it's:
(x) + (x+6) = 42

8 0

3 years ago

5. 18 m 11 m Not drawn to scale 29 m b. 445 m 7 m d. 203 m

Luden [163]

Answer: D

Explanation: Apply the pythagorean theorem (a^2 + b^2 = c^2) to get 11^2 + b^2 = 18^2. From that you can solve for b and you get the square root of 203.

Hope this helped! :)

8 0

3 years ago

Read 2 more answers

a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6?6x7x7=294 b) How many three-digit numbers

love history [14]

Answer:

a) 294

b) 180

c) 75

d) 168

e) 105

Step-by-step explanation:

Given the numbers 0, 1, 2, 3, 4, 5 and 6.

Part A)

How many 3 digit numbers can be formed ?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For unit's place, any of the numbers can be used i.e. 7 options.

For ten's place, any of the numbers can be used i.e. 7 options.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Total number of ways = $7 \times 7 \times 6$ = 294

Part B:

How many 3 digit numbers can be formed if repetition not allowed?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Now, one digit used, So For unit's place, any of the numbers can be used i.e. 6 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = $6 \times 6 \times 5$ = 180

Part C)

How many odd numbers if each digit used only once ?

Solution:

For a number to be odd, the last digit must be odd i.e. unit's place can have only one of the digits from 1, 3 and 5.

Number of options for unit's place = 3

Now, one digit used and 0 can not be at hundred's place So For hundred's place, any of the numbers can be used i.e. 5 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = $3 \times 5 \times 5$ = 75

Part d)

How many numbers greater than 330 ?

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 7

Number of options for unit's place = 7

Total number of ways = $3 \times 7 \times 7$ = 147

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 7

Total number of ways = $1 \times 3 \times 7$ = 21

Total number of required ways = 147 + 21 = 168

Part e)

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 6

Number of options for unit's place = 5

Total number of ways = $3 \times 6 \times 5$ = 90

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 5

Total number of ways = $1 \times 3 \times 5$ = 15

Total number of required ways = 90 + 15 = 105

7 0

3 years ago