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3 years ago

Answer pleaseeeeeeee

Mathematics

1 answer:

allsm [11]3 years ago

4 0

0.8 is the greatest and 1/9 is the least

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Ms hadley has an orange tree when she lived in california .less than 39 of the oranges were lost during the first month i need h

n200080 [17]

Answer:

Step-by-step explanation:boom boom boom boom i want you in my room lets spend the night together

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3 years ago

It costs Doug $30 to set up a cupcake stand. He charges $2 for every cupcake he sells. If Doug makes $48 of profit in one day, h

Lunna [17]

Answer:

It is 39 cupcakes.

Since he charges $2 per cupcake (39), it is 78.

He makes profit so subtract 30 from 78 and you get $48.

6 0

4 years ago

now has a book that is 3.6 meters long he cuts the Rope into two pieces one piece is 1.925 meters long how long is the other pie

Kamila [148]

The answer is 1.675 because when u subtract 3.6 from 1.925 it gives you 1.675

6 0

3 years ago

Read 2 more answers

Help me answer 99 plz

Vitek1552 [10]

One tick behind one is 3/4, 1/8 is basically half of 1/4, so find 1/4 and find 1/8, 0 is sort of obvious but then not obvious. three ticks behind 3/4. 2 2/3, count UP from 1 1/4 to find 2 2/3. I really hope I helped! :)

5 0

4 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