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loris [4]
3 years ago
14

A bakery offers a sale price of $3.45 for 4 muffins. What is the price per dozen?

Mathematics
1 answer:
Vlad [161]3 years ago
3 0

Answer:

Cost of 4 muffins = $3.45

Cost of 12 muffins = (3 sets of  x 4 muffins = 12)

Cost of 3 sets of muffins = $3.45 x3 = $10.35

Step-by-step explanation:

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Consider the quadratic equation x^2=4x-5. How many solutions does the equation have
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There is no solution

8 0
4 years ago
Some one please help!
nasty-shy [4]

Answer:

The answers are 23, 9,200 and 26, 10,400

Step-by-step explanation:


7 0
3 years ago
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How do I do this? Please help me
AnnyKZ [126]
You divide 46 by 5.2 and the answer should be 8.8 inches.
7 0
3 years ago
A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.
oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

H_0: \mu=2.4\\\\H_a:\mu> 2.4

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58

The degrees of freedom for this sample size are:

df=n-1=45-1=44

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>2.5801)=0.0066

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0
3 years ago
Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all
posledela

This question is Incomplete

Complete Question

Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all have the same body size in their adult phase, which made it easy to measure speeds in units of body lengths per second (bl/s). The researchers found that, when traffic is light and not congested, ant speeds vary roughly Normally, with mean 6.20 bl/s and standard deviation 1.58 bl/s. (a) What is the probability that an ant's speed in light traffic is faster than 5 bl/s? You may find Table B useful. (Enter your answer rounded to four decimal places.)

Answer:

0.7762

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

Population mean = 6.20 bl/s

Standard deviation = 1.58 bl/s.

x = 5 bl/s

z = 5 - 6.20/1.58

z = -0.75949

The probability that an ant's speed in light traffic is faster than 5 bl/s is P( x > 5)

Probability value from Z-Table:

P(x<5) = 0.22378

P(x>5) = 1 - P(x<5)

= 1 - 22378

= 0.77622

Approximately to 4 decimal places = 0.7762

The probability that an ant's speed in light traffic is faster than 5 bl/s is 0.7762

5 0
4 years ago
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