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4 years ago

If Marianne's father calls and says he needs 13 pounds of potting soil how many additional bags should she buy

Mathematics

1 answer:

Semenov [28]4 years ago

5 0

13 but it depends on how many pounds of poting soil the bags can carry

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If (44)x = 432, what is the value of x? 04 07 O 8 O 28

sdas [7]

Answer: x = 9.8181......

Step-by-step explanation:

(44)x =432

Take out Parentheses:

44x = 432

Divide both sides by 44:

x = 9.8181......

4 0

4 years ago

Which is an example of the commutative property?

astra-53 [7]

A. –13 + 27 = 27 + (–13) is an example of the commutative property.

The commutative property states that a + b = b + a.

8 0

3 years ago

Hellp me solve this thanks

stellarik [79]

Answer:

c. 10 cm

Step-by-step explanation:

The parallel sides of trapizium are the bases so

base 1(B1) =9cm , base(B2)= 12cm

$area(a) = 105 {cm}^{2}$

the perpendicular distance means the height so lets find the height

area of trapizium= B1 + B2 × h

2

${105cm}^{2}(the \: given \: area) = \frac{9cm(the \: given \: base) + 12cm(the \: given \: base)}{2} \times h$

${105cm}^{2} = \frac{21cm}{2} \times h$

now we crisscross

${105cm}^{2} \times 2 = 21cm \times h$

${210cm}^{2} = 21cm \times h$

$\frac{ {210cm}^{2} }{21cm} = \frac{21cm \: }{21cm} \times h$

h = 10 cm

hope u like it ‍❤️‍‍❤️‍‍❤️‍

For any question comment me

4 0

3 years ago

Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey interviews an SRS of 267 colleg

Fofino [41]

Answer:

The probability that 75% or more of the women in the sample have been on a diet is 0.037.

Step-by-step explanation:

Let X = number of college women on a diet.

The probability of a woman being on diet is, P (X) = p = 0.70.

The sample of women selected is, n = 267.

The random variable thus follows a Binomial distribution with parameters n = 267 and p = 0.70.

As the sample size is large (n > 30), according to the Central limit theorem the sampling distribution of sample proportions ( $\hat p$ ) follows a Normal distribution.

The mean of this distribution is:

$\mu_{\hat p} = p = 0.70$

The standard deviation of this distribution is: $\sigma_{\hat p}=\sqrt{\frac{p(1-p}{n}} =\sqrt{\frac{0.70(1-0.70}{267}}=0.028$

Compute the probability that 75% or more of the women in the sample have been on a diet as follows:

$P(\hat p \geq 0.75)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}} \geq \frac{0.75-0.70}{0.028}) =P(Z\geq0.179)=1-P(Z$

**Use the z-table for the probability.

$P(\hat p \geq 0.75)=1-P(Z$

Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.

7 0

3 years ago

A function is rule that assigns each

Alex_Xolod [135]

Answer:

element in the range to exactly one element in the domain

Step-by-step explanation:

N/A

6 0

3 years ago